Explanation:
Okay, so there's this gecko in a room, and I need to figure out the shortest path it can take to get from one point to another without jumping. The room is 12 feet long, 10 feet wide, and 8 feet tall. Let me try to visualize this. The gecko is on a side wall, which is 10 feet by 8 feet. Wait, the room has length, width, and height. Let me confirm: usually, in a room, length, width, and height. But the problem mentions a side wall that's 10 by 8. Wait, the room is 12 feet long, 10 feet wide, and 8 feet tall. So the walls would be:

- The front and back walls: length x height = 12 ft by 8 ft
- The left and right walls: width x height = 10 ft by 8 ft
- The floor and ceiling: length x width = 12 ft by 10 ft

So the gecko is on a side wall (which is width x height, so 10 by 8). The gecko is one foot from the ceiling and one foot from the back wall. Wait, the back wall is 12 by 8? Or is the back wall part of the length? Let me think again.

The problem says: the gecko is on a side wall (10' by 8'), one foot from the ceiling and one foot from the back wall (12' by 8'). So the side wall is 10 by 8, meaning that it's either the left or right wall. The back wall is 12 by 8, which would be the back wall (assuming that the front wall is also 12 by 8, and the left and right are 10 by 8). So the gecko is on a side wall (left or right, 10 by 8), one foot from the ceiling (so 8 - 1 = 7 feet from the floor?) and one foot from the back wall. Wait, the back wall is 12 by 8, so if the gecko is one foot from the back wall, that would be along the length of the room? Let me try to sketch mentally.

Let me consider the room as a rectangular box with length 12, width 10, height 8. Let me assign coordinates. Let me think of the room with coordinates (x, y, z), where x is along the length (from front to back), y is along the width (left to right), and z is height (floor to ceiling). So front wall is at x=0, back wall is at x=12, left wall is at y=0, right wall is at y=10, floor at z=0, ceiling at z=8.

So the gecko is on a side wall (which is y=0 or y=10). The problem says "one foot from the ceiling and one foot from the back wall". So if the gecko is on a side wall, which is y=0 or y=10. Let me assume it's on the left side wall (y=0). Then "one foot from the back wall" would mean along the x-axis. Since back wall is at x=12, so one foot from back wall would be at x=12 - 1 = 11 feet. Also, "one foot from the ceiling" would be z=8 - 1 = 7 feet. So the gecko is at position (11, 0, 7). Wait, but is that correct?

Wait, if the gecko is on the left wall (y=0), then the coordinates would be (x, 0, z). The back wall is at x=12, so one foot from back wall is x=12 - 1 = 11. And one foot from ceiling is z=8 - 1 = 7. So yes, (11, 0, 7).

Then the fly is on the opposite side wall, which would be the right wall (y=10). The fly is one foot from the floor and one foot from the front wall. Front wall is at x=0, so one foot from front wall is x=0 + 1 = 1 foot. One foot from floor is z=0 + 1 = 1 foot. Therefore, the fly is at (1, 10, 1).

So now, the gecko is at (11, 0, 7), and the fly is at (1, 10, 1). The gecko can walk on the ceiling and walls, but not jump. So the question is to find the shortest path from (11, 0, 7) to (1, 10, 1) along the surfaces of the room.

This seems like a classic shortest path on a surface problem, which is often solved by unfolding the room into a net, and then finding the straight line distance on the net.

So, to figure out the shortest path, we need to figure out which walls or ceiling the gecko will traverse. Since the gecko is on the left wall, and the fly is on the right wall, the shortest path might go across the ceiling or the floor or some combination of walls.

But since the gecko can walk on the ceiling and walls, perhaps the shortest path would involve unfolding the walls into a plane. Let me think about different possible unfoldings.

First, let me note that the gecko is on the left wall (y=0), and the fly is on the right wall (y=10). The straight-line distance through the room would be sqrt[(12 - 2)^2 + (10 - 0)^2 + (8 - 6)^2]? Wait, no, that's not correct because the gecko can't go through the room, it has to stay on the surfaces.

So, the key idea is to find a way to unfold the walls such that the path becomes a straight line. There are multiple ways to unfold the room into a plane, and we need to find the one that gives the shortest distance.

Let me try to think step by step.

First, the gecko is on the left wall (y=0). Let me imagine unfolding the walls adjacent to the left wall (the floor, ceiling, and front/back walls) into a plane. Similarly, for the right wall.

But since the gecko needs to reach the opposite side wall, maybe the path goes from the left wall to the ceiling or floor, then to the right wall? Or maybe across the front or back wall?

Alternatively, perhaps the gecko can go from the left wall, cross the ceiling to the right wall, or cross the floor to the right wall. Let me try to calculate both possibilities.

First, let's think about going from the left wall to the ceiling, then to the right wall. Alternatively, maybe going from the left wall to the floor then to the right wall.

Alternatively, the gecko could go from the left wall to the front wall, then to the right wall, or to the back wall?

But since the gecko is one foot away from the back wall (at x=11), and the fly is one foot away from the front wall (at x=1), maybe the path goes through the front wall?

But let me think. Let me try to visualize unfolding the walls.

Let me first think about the possibility of unfolding the left wall and the ceiling into a flat surface. So if we unfold the left wall (y=0) and the ceiling (z=8), how would that look? If we imagine the left wall (which is 10ft wide (y-axis) by 8ft tall (z-axis)) and the ceiling (which is 12ft long (x-axis) by 10ft wide (y-axis)). But how do they connect? The ceiling connects to the left wall along the top edge of the left wall (z=8, y=0). If we unfold the left wall and the ceiling, we can imagine rotating the ceiling so that it's adjacent to the left wall. But maybe there's a better way.

Alternatively, maybe if we unfold the left wall, the back wall, and the ceiling? But maybe I need to consider different combinations.

Alternatively, think of the room as a box, and the shortest path will cross two adjacent faces. For example, from the left wall to the ceiling, then to the right wall. Or from the left wall to the floor, then to the right wall. Or from the left wall to the front wall, then to the right wall, etc.

Let me try to consider different cases.

Case 1: Gecko goes from left wall to ceiling to right wall.

To calculate this, we need to imagine unfolding the left wall and the ceiling into a plane. So, if we take the left wall (y=0) and unfold the ceiling (z=8) adjacent to it. The left wall is 10ft (width) by 8ft (height). The ceiling is 12ft (length) by 10ft (width). If we unfold the ceiling upwards from the left wall, then the total surface would be a rectangle with length equal to the height of the left wall (8ft) plus the length of the ceiling? Wait, no, maybe.

Wait, when you unfold the left wall (y=0) and the ceiling (z=8), the left wall is vertical, and the ceiling is horizontal. If we unfold them into a plane, the left wall would be connected along the top edge to the ceiling. So if you imagine flattening the ceiling so that it becomes adjacent to the left wall, then the combined surface would be a rectangle. Let me think of coordinates. On the left wall, the gecko is at (x=11, y=0, z=7). Wait, but in terms of the left wall, which is y=0, x ranges from 0 to 12, z from 0 to 8.

But when we unfold the left wall and the ceiling, the ceiling is attached along the top edge (z=8) of the left wall. So if we unfold the ceiling by rotating it 90 degrees around the top edge, then the ceiling would be adjacent to the left wall. So in this case, the left wall and the ceiling form a larger rectangle. The left wall has dimensions 12 (x) by 8 (z). The ceiling has dimensions 12 (x) by 10 (y). But when unfolded, how do they connect? Wait, maybe I need to think of the left wall's y-axis? Wait, no, the left wall is at y=0, so its coordinates are x and z. The ceiling is at z=8, and its coordinates are x and y. So when you unfold them, you need to align the edge where they meet.

The edge where the left wall meets the ceiling is at y=0, z=8. So if you unfold the ceiling so that it's adjacent to the left wall, you can imagine the ceiling extending from the left wall. However, since the ceiling is 10 feet wide (y-direction) and 12 feet long (x-direction), but the left wall is 12 feet long (x-direction) and 8 feet tall (z-direction). So if you unfold the ceiling so that it's adjacent to the left wall along the top edge (z=8, y=0), then the ceiling would be folded outwards from the left wall. Therefore, the combined surface would be a rectangle that is 12 feet (x) in length, and 8 (z) + 10 (y)? Wait, but that might not be right. Wait, when you unfold the ceiling, you can think of it as extending in the y-direction from the left wall's edge. Wait, maybe it's better to think of the left wall and the ceiling as forming a larger plane.

Alternatively, perhaps if you unfold the left wall and the ceiling into a plane, the total distance would be calculated by considering the gecko moving from its position on the left wall to some point on the ceiling, then to the fly's position on the right wall. But this seems complex. Maybe it's better to use the net method.

Alternatively, think of the path as crossing two adjacent faces: the left wall and the ceiling. If I can represent those two faces as a single plane, then the straight-line distance would be the shortest path.

Similarly, if the gecko crosses the left wall and the front wall, or left wall and the floor, etc.

Alternatively, think of unfolding the room in such a way that the gecko's path becomes a straight line. Since the gecko is on the left wall, and the fly is on the right wall, maybe the path crosses the ceiling or the floor.

Let me consider unfolding the left wall and the ceiling. Let me imagine the left wall (size 12x8) and the ceiling (12x10). When unfolded, the left wall and ceiling are adjacent. But how?

Wait, the left wall is at y=0, with x from 0 to 12, z from 0 to 8. The ceiling is at z=8, with x from 0 to 12, y from 0 to 10. So if you unfold the ceiling so that it's attached to the left wall's top edge (z=8), then the ceiling would be attached along the top of the left wall. Therefore, if you were to lay the ceiling flat next to the left wall, the combined surface would be a rectangle with dimensions x from 0 to 12, and z from 0 to 8 (left wall) plus y from 0 to 10 (ceiling)? Wait, maybe not. Wait, if we consider the left wall as a vertical rectangle and the ceiling as a horizontal rectangle attached to its top, then in the unfolded net, the vertical left wall and the horizontal ceiling would form a larger rectangle with dimensions length in x-direction 12, and height in z and y directions? Wait, this is getting confusing.

Maybe it's easier to think of coordinates. Suppose we unfold the left wall and the ceiling into a plane. Let me assign coordinates to each face.

Original coordinates:

Left wall (y=0): x (from 0 to 12), z (from 0 to 8).

Ceiling (z=8): x (from 0 to 12), y (from 0 to 10).

When we unfold the ceiling so that it's adjacent to the left wall, the ceiling is now in the same plane as the left wall. Let me imagine "flattening" the ceiling so that it's next to the left wall. So, if you take the ceiling and fold it down onto the left wall, but actually, to make it a plane, it's like taping the ceiling to the left wall along the top edge. However, in terms of coordinates, the ceiling's y-coordinate is perpendicular to the left wall's z-coordinate. To flatten them, we need to represent them as a single plane.

Alternatively, think of the left wall as having coordinates (x, z) where 0 ≤ x ≤ 12, 0 ≤ z ≤ 8. When unfolded with the ceiling, which has coordinates (x, y) where 0 ≤ x ≤ 12, 0 ≤ y ≤ 10. To represent them on the same plane, we can imagine that the ceiling is now in the same plane as the left wall, but with y and z coordinates combined. Wait, maybe not. Alternatively, we can consider reflecting coordinates.

Alternatively, think of the ceiling as extending from the left wall's top edge. So, if we consider moving from the left wall to the ceiling, the path would cross from (x, z) on the left wall to (x, y) on the ceiling, but since they are adjacent, we can represent this as a single plane where the ceiling is "folded" over the left wall.

But this is abstract. Maybe I can use the method of reflecting the room or considering the net as a combination of walls.

Wait, here's a better method: to find the shortest path on the surface, imagine unfolding the room so that the start and end points lie on the same plane, and then the straight line distance on that plane is the shortest path.

So, for example, if the gecko goes from left wall to ceiling to right wall, we can imagine unfolding the left wall and the ceiling into a single plane, and then the fly is on the right wall. But the right wall is not directly connected to the ceiling or left wall. Wait, the right wall is connected to the ceiling. So if we unfold the ceiling, then the right wall is adjacent to the ceiling.

Alternatively, perhaps the gecko goes from the left wall across the ceiling to the right wall. So, to represent this path as a straight line, we need to unfold the left wall and the ceiling and the right wall into a single plane. Wait, but how?

Alternatively, think of the left wall, ceiling, and right wall as a strip. If we unfold the left wall (y=0), the ceiling (z=8), and the right wall (y=10) all in one plane. But the ceiling is connected to both the left and right walls. If we lay the left wall, ceiling, and right wall flat, they would form a larger rectangle where the left wall is connected to the ceiling, and the ceiling is connected to the right wall. Therefore, the total length in the x-direction would be the same, but the height would be the height of the wall (8ft) plus the width of the ceiling (10ft)? Wait, no.

Wait, if you unfold the left wall, ceiling, and right wall, you can imagine the left wall (12x8), then the ceiling (12x10), then the right wall (12x8), connected along the x-axis. Wait, but the left wall and right wall are both connected to the ceiling. So when unfolded, they would form a sort of L-shape? No, if you lay them flat, they would be a straight line? Wait, no. If you take the left wall and unfold the ceiling so that it's adjacent to the left wall, then unfold the right wall adjacent to the ceiling. So, the left wall, ceiling, right wall would form a sort of long rectangle?

Wait, let me think of the left wall as a vertical rectangle, then the ceiling is horizontal, attached to the top of the left wall, and then the right wall is vertical, attached to the end of the ceiling. But actually, the right wall is also vertical, but connected to the ceiling. If you unfold the ceiling between the left and right walls, then the left wall, ceiling, right wall would form a flat plane where the total dimensions are?

In terms of length along x-axis: left wall is 12 feet (x from 0 to 12), the ceiling is 12 feet (x from 0 to 12), and the right wall is 12 feet (x from 0 to 12). But if we combine them, it's more about the other dimensions.

Alternatively, think of the left wall as having width in the z-direction (height) and the ceiling as having width in the y-direction. If we unfold them, we can represent them as a single plane with coordinates combining z and y. Wait, this is getting too abstract. Maybe it's better to use coordinates.

Suppose we imagine the gecko's starting point (11, 0, 7) on the left wall. If we unfold the left wall and the ceiling into a plane, then on this plane, the coordinates of the gecko would be (x, z) where z is height, and then the ceiling is unfolded so that it's adjacent. If we want to go from the left wall to the ceiling to the right wall, we need to consider coordinates on both.

Alternatively, another approach is to consider that when moving across two adjacent faces, the path will be a straight line in the unfolded net. Therefore, to calculate the distance, we can calculate the straight line between the start and end points on the net.

But to do that, we need to figure out which faces are involved in the shortest path. Since the gecko is on the left wall and the fly is on the right wall, the shortest path may go across the ceiling or the floor, or across the front/back wall.

Let me try considering different possibilities:

Possibility 1: The gecko goes from left wall to ceiling to right wall.

Possibility 2: The gecko goes from left wall to floor to right wall.

Possibility 3: The gecko goes from left wall to front wall to right wall.

Possibility 4: The gecko goes from left wall to back wall to right wall.

Possibility 5: The gecko goes across multiple walls, like left wall to ceiling to front wall to right wall, etc. But since we want the shortest path, it's likely that it goes through two walls.

So let's evaluate Possibility 1 and 2 first.

Starting with Possibility 1: Left wall to ceiling to right wall.

If we unfold the left wall and the ceiling so that they are on the same plane, then the gecko's starting point is on the left wall, and the fly's position on the right wall is now on the ceiling's side? Wait, not sure.

Wait, to make this concrete, let's think of the left wall as one face and the ceiling as another face. Let me imagine unfolding the left wall (y=0) and the ceiling (z=8) into a single plane. If we do that, then the gecko's position on the left wall is still at (x=11, z=7). However, the fly is on the right wall (y=10), which is not directly connected to the ceiling unless we also unfold the right wall. Therefore, to have the fly on the same plane as the gecko's path, we need to unfold the ceiling and the right wall as well.

Alternatively, think about unfolding the left wall, ceiling, and right wall into a single plane. If we do that, then the left wall, ceiling, and right wall form a sort of 3D shape that can be flattened into a 2D plane. The left wall is connected to the ceiling, and the ceiling is connected to the right wall. When unfolded, this would create a flat surface where the left wall and right wall are on opposite ends, connected via the ceiling.

In such a case, the total width (in the direction perpendicular to the x-axis) would be the height of the left wall (8 feet) plus the width of the ceiling (10 feet)? Wait, no. Wait, if we lay the left wall, ceiling, and right wall flat, the dimensions would be?

Wait, the left wall has dimensions 12 feet (x-direction) by 8 feet (z-direction). The ceiling has 12 feet (x-direction) by 10 feet (y-direction). The right wall has 12 feet (x-direction) by 8 feet (z-direction). When unfolded, the ceiling is between the left and right walls? If we lay them out so that the left wall is adjacent to the ceiling along the top edge (z=8), and the ceiling is adjacent to the right wall along its side (y=10). Therefore, if you lay this all flat, the total height in the y-z direction would be the height of the left wall (8ft) plus the width of the ceiling (10ft)? Wait, not quite. Wait, when unfolded, the left wall and ceiling are in one plane, and the ceiling and right wall are in another? No, it's all in one plane. Hmm.

Alternatively, think of the left wall as a vertical rectangle, and then the ceiling is like a horizontal rectangle connected to the top of the left wall. Then the right wall is connected to the end of the ceiling. But in terms of coordinates, if we create a net where the left wall is on the left, the ceiling is above it, and the right wall is to the right of the ceiling, then the entire net would be a sort of T-shape? Wait, maybe not. Alternatively, imagine the left wall and ceiling forming a vertical rectangle with the ceiling folded out. If we then fold the right wall out from the ceiling, it becomes a different orientation.

This is getting too vague. Let me try using coordinates with unfolding.

Let me consider unfolding the left wall (y=0) with the ceiling (z=8). When unfolded, the ceiling is adjacent to the left wall. So, the left wall's top edge (z=8) is connected to the ceiling's bottom edge (z=8). Therefore, if we were to represent this as a single plane, we can think of the left wall as going from z=0 to z=8, and the ceiling extending in the y-direction from y=0 to y=10. Wait, but in terms of coordinates, if we unfold the ceiling so that it is in the same plane as the left wall, then the coordinates can be combined.

Alternatively, if we think of the left wall's z-coordinate as the vertical axis and the ceiling's y-coordinate as the horizontal axis. Wait, maybe not. Let me try to assign coordinates.

Let me denote that on the left wall (y=0), the gecko is at position (x=11, z=7). The fly is on the right wall (y=10) at (x=1, z=1). If we unfold the left wall and the ceiling, the ceiling is adjacent to the left wall. Let me imagine the ceiling being laid flat so that it's adjacent to the left wall's top edge. So the ceiling is now extending in the y-direction from the left wall. Therefore, the coordinates on the unfolded net would be:

- On the left wall: (x, z) where 0 ≤ x ≤ 12, 0 ≤ z ≤ 8.

- On the ceiling: (x, y) where 0 ≤ x ≤ 12, 0 ≤ y ≤ 10, but in the unfolded net, this is adjacent to the left wall's top edge. So if we consider the left wall as having z from 0 to 8, the ceiling would be in a different dimension. Maybe we can map the ceiling's y to z? Wait, if we unfold the ceiling down from the left wall, then the ceiling's y becomes an extension in the z-direction? Not sure. Maybe this is getting too confusing.

Alternatively, think of the path as going from the gecko's starting point on the left wall, across the ceiling, to the fly's position on the right wall. To compute the distance, imagine that the ceiling is a rectangle (12x10) and the left and right walls are connected to it.

But how to model this? Perhaps, if we consider the three-dimensional coordinates, but when unfolded, the ceiling is flat. Let me think of the gecko's path as moving from (11, 0, 7) on the left wall to some point on the ceiling, then to (1, 10, 1) on the right wall. However, to find the shortest path, it's equivalent to finding a straight line in the unfolded net.

Alternatively, think of the ceiling as a bridge between the left and right walls. If we unfold the left wall, ceiling, and right wall into a single plane, the total horizontal space (in the x-direction) is still 12 feet, but the vertical space would be the height of the left wall (8 ft) plus the width of the ceiling (10 ft)? Wait, no, the width of the ceiling is in the y-direction, which is perpendicular to the x-direction. This is getting too tangled.

Wait, here's a different approach. The problem is similar to finding the shortest path on the surface of a rectangular prism between two points. The standard method is to consider different possible unfoldings and calculate the distance for each, then take the minimum.

In order to do that, I need to figure out which faces the gecko would traverse. Since the gecko is on the left wall and the fly is on the right wall, the shortest path will go through two adjacent faces. The two faces could be:

1. Left wall and ceiling (path goes from left wall to ceiling to right wall)
2. Left wall and floor (path goes from left wall to floor to right wall)
3. Left wall and front wall (path goes from left wall to front wall to right wall)
4. Left wall and back wall (path goes from left wall to back wall to right wall)

Also, maybe the path goes through three faces, but probably not necessary because the shortest path would cross two faces.

Let me calculate the distances for each of these cases.

Case 1: Left wall to ceiling to right wall.

To model this, imagine unfolding the left wall and the ceiling into a single plane. Then the fly would be on the right wall, which is connected to the ceiling. Wait, so maybe we need to include the right wall as well?

Alternatively, think of the left wall, ceiling, and right wall as a single plane. When unfolded, the left wall is connected to the ceiling, and the ceiling is connected to the right wall. Therefore, the total dimensions would be the x-direction is still 12 feet (from front to back), but the other dimension is the height of the left wall (8 feet) plus the width of the ceiling (10 feet)? Wait, no. Wait, if we take the left wall (which is 12x8), the ceiling (12x10), and the right wall (12x8), when unfolded, they form a shape that's 12 feet in length, and the height is 8 + 10 = 18 feet? Wait, but the right wall is connected to the back of the ceiling. Wait, no, the ceiling is in the middle.

Wait, maybe if we unfold the left wall, the ceiling, and the right wall into a single plane, the total height would be the height of the left wall (8) plus the width of the ceiling (10) plus the height of the right wall (8)? No, because the right wall is connected to the ceiling's side. Let me try to visualize.

Suppose we have the left wall, then the ceiling, then the right wall. If you unfold them so that they are all in the same plane, the left wall is vertical, the ceiling is horizontal, and the right wall is vertical again. If you lay them out flat, the total "height" would be the height of the left wall (8) plus the width of the ceiling (10) plus the height of the right wall (8)? That seems too much. Wait, no. When you unfold them, the ceiling is between the left wall and the right wall. So, if you have the left wall, then the ceiling, then the right wall, they are connected along the x-axis. Therefore, the total "width" in the y-z plane would be?

This is really tricky without a diagram. Maybe I need to use a different method.

Let me recall that in such problems, the shortest path can be found by reflecting the room. For example, reflecting the fly's position across the walls and finding the straight line.

Alternatively, think about the net. For instance, if the gecko goes from the left wall to the ceiling to the right wall, then we can imagine folding the left wall and the ceiling into a single plane, then the fly's position on the right wall is... Hmm. Wait, let me consider coordinates again.

Let me think of the left wall (y=0) and the ceiling (z=8). Let me consider the ceiling as extending from the left wall. If I unfold the left wall and ceiling into a plane, then the coordinates of the gecko on the left wall are (x, z), and on the ceiling, they would be (x, y) where y is the distance from the left wall. Wait, if the ceiling is unfolded adjacent to the left wall, then moving from the left wall to the ceiling would be moving in the y-direction. So, if the gecko walks from the left wall to the ceiling, then on the unfolded net, the gecko's path would be a straight line from its position on the left wall to some point on the ceiling, and then to the fly's position on the right wall.

But since the fly is on the right wall (y=10), which is connected to the ceiling, we need to figure out how to represent the fly's position on this unfolded net.

Alternatively, think of the entire path from the gecko's starting point to the fly's position as crossing the left wall, ceiling, and right wall. To represent this as a straight line, we need to have all three faces unfolded into a plane. However, due to the complexity, perhaps it's easier to use the method of reflecting the room.

Wait, here's a trick: if you imagine reflecting the room across the ceiling or floor, so that the gecko's path can be represented as a straight line across multiple reflections. For example, reflecting the fly's position across the ceiling to find a straight line path.

But since the gecko is allowed to walk on the ceiling and walls, the shortest path can be found by finding a straight line in the unfolded net where the gecko's start and end points lie on adjacent faces.

Alternatively, think about the following: If we unfold the left wall and the ceiling such that the right wall is adjacent to the ceiling, then the total distance can be calculated as follows:

The gecko starts at (11, 0, 7). If we unfold the left wall and the ceiling, the coordinates on the left wall are (x, z), and on the ceiling, they are (x, y). However, when we unfold the ceiling, so that it's adjacent to the left wall, we can imagine that the ceiling is now in the y-direction. So if we consider the ceiling unfolded, then the gecko can go from (x, z) to (x, y). But the fly is on the right wall (y=10). So perhaps if we unfold the ceiling and the right wall, then the fly is on the right wall, which is connected to the ceiling.

Alternatively, here's an idea: imagine that the gecko travels from the left wall to the ceiling to the right wall. If we unfold the left wall and the ceiling into a single plane, then the fly's position on the right wall can be represented as a point on the ceiling. Wait, how?

If we consider that the gecko travels from the left wall to the ceiling to the right wall, then on the unfolded net (left wall + ceiling), the fly's position is actually on the right wall, which is connected to the ceiling. So, if we imagine the right wall as being folded out from the ceiling, then on the unfolded net, the right wall would be adjacent to the ceiling, forming a larger plane.

Therefore, the coordinates on the unfolded net would be: for the left wall, (x, z), and on the ceiling and right wall, (x, y). Therefore, the total dimensions are: in the x-direction, it's 12 feet, and in the y-z combined direction? But how to map this.

Alternatively, think of it as follows: when unfolding the left wall, ceiling, and right wall into a single plane, the total width in one direction is the length of the room (12 feet), and the other direction is the height of the left/right wall (8 feet) plus the width of the ceiling (10 feet). Therefore, the total distance would be calculated as a straight line across this combined width.

But how exactly?

Wait, suppose we have the left wall (12x8) and the ceiling (12x10) and the right wall (12x8). If we unfold them so that they are all in a straight line along the y-direction, then the total length would be 8 (height of left wall) + 10 (width of ceiling) + 8 (height of right wall) = 26 feet? But that seems too long. Wait, no. If you lay them out side by side, the total length would be the sum of their widths in the respective directions. This is confusing.

Let me try to think of it in terms of coordinates again. If the gecko is on the left wall at (11, 0, 7), and the fly is on the right wall at (1, 10, 1). Suppose we unfold the left wall and the ceiling so that they form a plane. Then, the gecko's position on the left wall can be represented as (x, z), and the fly's position on the right wall is on the ceiling. Wait, no, the fly is on the right wall, not the ceiling.

Alternatively, if we go from the left wall to the ceiling, then to the right wall, then we can consider the ceiling as a bridge between them. So, on the ceiling, the fly is at (x=1, z=1), but since it's on the right wall, which is y=10, how does that translate?

Wait, if the gecko is on the left wall at (11, 0, 7), and it goes to the ceiling, which is at z=8. The ceiling is at z=8, and the fly is on the right wall at (1, 10, 1). So if we consider the gecko crossing from the left wall to the ceiling, then from the ceiling to the right wall, the coordinates on the ceiling would be (x, y). So, when on the ceiling, the gecko's path would be from (x1, y1) to (x2, y2). But since the gecko is moving from the left wall to the ceiling, which is adjacent, we can imagine that the starting point on the left wall corresponds to a point on the ceiling.

Alternatively, think of the left wall and ceiling as a single plane where the ceiling is the extension of the left wall's top. So, for example, if we have the left wall's top edge (z=8) connected to the ceiling's bottom edge (z=8). Then, if we imagine "flattening" the ceiling so that it's on the same plane as the left wall, then the coordinates on the ceiling would be (x, y) where y is measured from the left wall. Wait, that might work.

Therefore, on the left wall, the gecko is at (x=11, z=7). If we unfold the ceiling to be adjacent, we can represent the gecko's position as (x=11, z=7), and the fly is on the right wall (y=10). To reach the fly, the gecko would need to go from (11, 7) on the left wall to some point on the ceiling, then to the fly's position on the right wall. But since the fly is on the right wall, which is connected to the ceiling, we need to represent that on the unfolded net.

If we imagine that the ceiling is now extending in the y-direction from the left wall's top edge, then the fly's position on the right wall (which is at y=10) would be at a distance of y=10 from the left wall. However, since the ceiling is at y=0 to y=10, and connected to the left wall, the fly's position is at y=10 on the right wall. Wait, but the right wall is also at y=10. If we have the ceiling unfolded adjacent to the left wall, then the right wall is not directly connected. So perhaps we need to also unfold the right wall.

This is getting too complicated. Maybe I need to use the method of reflecting points across walls.

For example, to find the shortest path that goes from the left wall to the ceiling to the right wall, we can reflect the fly's position across the ceiling to the other side, and then find the straight line from the gecko's position to the reflection.

Wait, here's a method I remember: for problems involving shortest paths on the surface of a rectangular prism, you can reflect the room across the faces and find the straight-line distance between the original point and the reflected point.

For instance, if the gecko is moving from the left wall to the ceiling to the right wall, we can reflect the fly's position over the ceiling, but I need to think carefully.

Alternatively, consider that when moving from the left wall to the ceiling to the right wall, it's equivalent to going through two faces. So, imagine unfolding those two faces into a plane. For example, if we unfold the left wall and the ceiling, then the fly's position is on the right wall, which is a different face. Therefore, we might need to reflect the fly's position across the ceiling to a new position, and then calculate the distance from the gecko's original position to this reflection.

But I'm not sure. Let me try to find an example.

Suppose the gecko is on the left wall at (11, 0, 7), and the fly is on the right wall at (1, 10, 1). Let me consider unfolding the left wall and the ceiling. When we unfold them, the coordinates on the ceiling would be such that moving from the left wall to the ceiling is like moving along the y-axis. However, since the fly is on the right wall, which is at y=10, which is the end of the ceiling's width (y=10), maybe we can think of the fly's position on the ceiling as (1, 10, 1)? But no, because the ceiling is at z=8, so the fly is on the right wall, not the ceiling.

This is really challenging. Maybe I should look for another approach.

Let me think of the room as a rectangular prism with length=12, width=10, height=8. The gecko is at (11, 0, 7) on the left wall, and the fly is at (1, 10, 1) on the right wall. I need to find the shortest path along the surfaces.

One way to do this is to consider different possible unfoldings and calculate the distance. Let's try to think of possible unfoldings:

Unfolding 1: Left wall + ceiling + right wall. If we unfold the left wall, ceiling, and right wall so that they form a flat plane. This would create a large rectangle. Let's imagine that the left wall is on the left, the ceiling in the middle, and the right wall on the right. When unfolded, the total horizontal length is the same as the room's length (12 feet), and the vertical dimension is the height of the left wall (8 feet) plus the width of the ceiling (10 feet) plus the height of the right wall (8 feet)? No, that might not be correct.

Actually, if you unfold the left wall, ceiling, and right wall in sequence, the vertical dimension would be the height of the left wall plus the width of the ceiling? No, since the ceiling is horizontal. Wait, let's think about the dimensions:

- The left wall has dimensions 12 (length) x 8 (height).
- The ceiling has dimensions 12 (length) x 10 (width).
- The right wall has dimensions 12 (length) x 8 (height).

When we unfold the left wall, ceiling, and right wall in sequence, the total area would be a combination of these. However, to flatten them, the dimensions would be:

If you lay the left wall, then the ceiling, then the right wall next to each other, the total length in the direction of the length of the room (x-axis) is still 12 feet. The width would be the height of the left wall (8) + width of ceiling (10) + height of right wall (8)? No, that would be if they are stacked vertically, but since the ceiling is a horizontal surface, it's more like attaching it to the left wall's top, and then attaching the right wall to the end of the ceiling. So the combined dimensions would be:

- In the x-direction: 12 feet (same as room length).
- In the y-direction: 8 feet (height of left wall) + 10 feet (width of ceiling) + 8 feet (height of right wall)? No, that seems wrong.

Wait, actually, when you attach the ceiling to the left wall's top edge, the ceiling extends in the y-direction (width of the room), and the right wall is attached to the ceiling's end in the y-direction. So, the total dimension in the y-direction would be the height of the left wall (8) plus the width of the ceiling (10)? Because the left wall is vertical (height) and the ceiling is horizontal (width). If we unfold them such that the left wall is on the left, and the ceiling extends to the right from the top of the left wall, then the right wall would be attached to the end of the ceiling.

In this case, the total vertical dimension from the left wall's bottom to the right wall's top would be 8 (left wall height) + 10 (ceiling width) + 8 (right wall height)? No, that doesn't make sense. Let me think of it as two dimensions. If we have the left wall (12x8) unfolded with the ceiling (12x10), and then the right wall (12x8) added, the combined shape would have dimensions of 12 feet (length) by (8 + 10 + 8) feet? No, because the ceiling is attached to the top of the left wall, so it's not adding to the same dimension. This is not working.

Maybe I need to use coordinates. If I unfold the left wall and the ceiling such that the ceiling is attached to the top of the left wall, then the gecko's starting position is on the left wall at (x=11, z=7). The fly is on the right wall at (x=1, y=10, z=1). To represent this on the unfolded net, I need to find a coordinate system where both the gecko's position and the fly's position can be represented.

If I unfold the left wall and the ceiling, then the ceiling is in the y-direction from y=0 to y=10, and the left wall is in the z-direction from z=0 to z=8. But how do these combine?

Maybe if we consider the ceiling as extending in the y-direction from the left wall, then any point on the ceiling can be represented as (x, y), where y is from 0 to 10, and x is from 0 to 12. Meanwhile, the left wall is (x, z) where z is from 0 to 8.

To find the shortest path from (11, z=7) on the left wall to (1, y=10, z=1) on the right wall, we can imagine that the path goes from the left wall to the ceiling to the right wall. To model this on the unfolded net, we could reflect the fly's position across the ceiling.

Wait, if we reflect the fly's position across the ceiling, but I need to think.

Alternatively, think of the ceiling as a bridge. The gecko goes from left wall to ceiling to right wall. The straight-line distance on this unfolded net would be the hypotenuse of a right triangle, where one leg is the horizontal distance between the left wall and right wall, and the other is the vertical distance.

Wait, the horizontal distance between the left wall and the right wall is the length of the room, 12 feet, but since the gecko is one foot from the back wall (x=11) and the fly is one foot from the front wall (x=1), the distance along the x-axis is 11 - 1 = 10 feet? Or is it 12 - 1 - 1 = 10 feet?

Wait, the gecko is at x=11 on the left wall, and the fly is at x=1 on the right wall. So the distance along the x-axis between them is 11 - 1 = 10 feet? But since they are on opposite sides, it's actually 12 - 1 - 1 = 10 feet? Yes, because from x=1 (front) to x=11 (back) is 10 feet.

However, the path goes across the ceiling, so the vertical distance would involve the gecko's height and the fly's height.

Wait, the gecko is at z=7 (one foot from ceiling), and the fly is at z=1 (one foot from floor). If the gecko crosses the ceiling, it would go from z=7 to z=8 (ceiling) and then to z=1 on the right wall. But how does this translate in terms of distance?

Alternatively, if we consider the vertical distance as the difference in heights plus the height of the ceiling? No, this is confusing.

Let me try to think of it as a 2D problem by unfolding the left wall and ceiling into a plane. Let's say we have the left wall (x from 0 to 12, z from 0 to 8) and the ceiling (x from 0 to 12, y from 0 to 10). If we unfold the ceiling so that it is adjacent to the top edge of the left wall, then the combined surface is a rectangle with length 12 feet (x-axis) and height 8 + 10 = 18 feet (z + y). Wait, is that correct?

Yes! If you imagine unfolding the ceiling so that it is attached to the top of the left wall, then the total vertical dimension becomes the height of the left wall (8 feet) plus the width of the ceiling (10 feet). So in this unfolded net, the gecko's starting position is at (x=11, z=7) on the left wall. The fly is on the right wall, which is at y=10, but how does that translate?

Wait, if the ceiling is unfolded, then the right wall is connected to the end of the ceiling. So, if we imagine the unfolded net as a rectangle of 12 (x) by 18 (z+y), then the right wall is at y=10, which is 10 feet from the left wall (y=0). However, in this unfolded net, the coordinates would be such that moving from the left wall to the ceiling is moving in the y-direction. Therefore, the fly's position on the right wall is at (x=1, y=10, z=1). Wait, but on the unfolded net, the right wall is part of the ceiling?

No, the right wall is a separate face. If we have unfolded the left wall and the ceiling, the right wall is still separate. So to include the right wall, we need to unfold it as well.

Alternatively, if we include the right wall in the unfolding, then the total unfolded net would be the left wall, ceiling, and right wall, forming a larger rectangle. The left wall has width 12 (x) and height 8 (z). The ceiling has width 12 (x) and height 10 (y). The right wall has width 12 (x) and height 8 (z). If we unfold them such that they are connected along the x-axis, the total height would be 8 (left wall) + 10 (ceiling) + 8 (right wall)? No, because the right wall is at y=10, which is the same as the ceiling's width.

Actually, if you unfold the left wall, then the ceiling, then the right wall, all connected along the x-axis, the total height would be the height of the left wall (8) plus the width of the ceiling (10) plus the height of the right wall (8)? No, that would be if they are stacked vertically, but since the ceiling is horizontal, it's more like extending in the y-direction.

This is really too confusing without a diagram. Let me try to look for another approach.

Let me consider that the shortest path will cross two adjacent faces. For example, the left wall and the ceiling, or the left wall and the floor, etc. Let's calculate the distance for each case.

Case 1: Left wall to ceiling to right wall.

To calculate this, imagine that the gecko travels from its position on the left wall, across the ceiling, to the fly's position on the right wall. If we unfold the left wall and the ceiling into a single plane, the gecko's starting point is on the left wall, and the fly's position is on the right wall. However, since the right wall is connected to the ceiling, we need to adjust coordinates.

Let me define the unfolded net as follows: the left wall is a vertical rectangle, and the ceiling is a horizontal rectangle attached to the top of the left wall. In this net, the gecko's position is at (x=11, z=7), and the fly's position is on the right wall, which is at the end of the ceiling. But how to represent the fly's position?

The fly is on the right wall at (x=1, y=10, z=1). On the unfolded net, the right wall is not directly connected to the ceiling; it's connected to the ceiling's end in the y-direction. So if we consider the ceiling as extending in the y-direction, then the fly's position on the right wall is at y=10, x=1, z=1. Wait, but in the unfolded net, if we have the left wall and ceiling, the right wall is not included. So to include the right wall, we need to unfold it too.

Alternatively, imagine that the ceiling is connected to the right wall, so the total unfolded net is left wall + ceiling + right wall. In this case, the gecko's position is on the left wall at (11, 7), and the fly's position is on the right wall at (1, 10, 1). To find the straight-line distance on this unfolded net, we need to represent both positions in the same coordinate system.

In this unfolded net, the left wall is a rectangle of 12 x 8, the ceiling is 12 x 10, and the right wall is 12 x 8. If we lay them out such that the left wall is on the left, the ceiling is above it, and the right wall is on the right of the ceiling, then the coordinates would be:

- For the left wall: x from 0 to 12, z from 0 to 8.
- For the ceiling: x from 0 to 12, y from 0 to 10 (but attached to the top of the left wall).
- For the right wall: x from 0 to 12, z from 0 to 8 (attached to the end of the ceiling in the y-direction).

But this is still unclear. Let me try to assign coordinates in this unfolded net.

Let me imagine that in this net, we have:

- The left wall is the vertical part from (0, 0) to (12, 8) in a coordinate system where x is the length (0-12) and y is the height (0-8).

- The ceiling is attached to the top of the left wall, extending in the y-direction from (0, 8) to (12, 8 + 10 = 18). So the ceiling is a horizontal strip from (0, 8) to (12, 18).

- The right wall is attached to the end of the ceiling, extending in the z-direction from (12, 8) to (12 + 8 = 20, 18)? No, this is getting too convoluted.

Alternatively, let me think of the combined unfolded net as having the left wall, ceiling, and right wall forming a rectangle with length 12 (x-axis) and height 8 + 10 = 18 (z + y). Wait, if I consider the left wall (z=0 to 8) and the ceiling (y=0 to 10), then if I unfold them, the total height would be 8 + 10 = 18. But the right wall is at y=10, so in this unfolded net, the fly's position is at (x=1, y=10, z=1)? No, I need to relate it.

Wait, if the gecko is on the left wall at (x=11, z=7), and in the unfolded net, the left wall is from x=0 to 12, z=0 to 8, then the gecko's position is at (11, 7). The fly is on the right wall at (x=1, y=10, z=1). But in the unfolded net, the right wall is at y=10, so if we consider the unfolded net as z from 0 to 8 (left wall) and y from 0 to 10 (ceiling), then the fly's position would be at (x=1, y=10, z=1)? But how does z=1 relate to this?

This is really too confusing. Maybe I need to use a different method.

Let me try using the method of unfolding the room such that the path crosses two adjacent faces. For example, if the gecko goes from the left wall to the ceiling to the right wall, then the total horizontal distance would be the distance between the gecko and the fly along the x-axis, and the total vertical distance would be the distance from the gecko's height to the fly's height plus the height of the ceiling?

Wait, the gecko is one foot from the ceiling (z=7), and the fly is one foot from the floor (z=1). If the gecko crosses the ceiling, it would go from z=7 to z=8 (ceiling) and then to z=1 on the right wall. So the vertical distance would be (8 - 7) + (8 - 1) = 1 + 7 = 8? Wait, no, that's if it goes up and down. But in terms of the straight line on the unfolded net, it would be the distance between the gecko's position and the fly's position, accounting for the unfolded dimensions.

Let me imagine that by unfolding the left wall and the ceiling, the gecko's position is at (11, 7) in a coordinate system where x is length and z is height, and the fly's position is somewhere on the ceiling. But the fly is on the right wall, not the ceiling, so perhaps I need to also include the right wall in the unfolding.

Alternatively, think of the gecko's path as going from (11, 0, 7) to (1, 10, 1). If we unfold the left wall and the ceiling, then the ceiling is 12 x 10, and the fly is on the right wall. If we also unfold the right wall, then the total dimensions would be such that the gecko's path is a straight line across the left wall, ceiling, and right wall.

But without knowing the exact unfolding, it's hard to proceed. Let me try to think of it as a 3D problem and project it onto 2D.

Let me consider the gecko's path as going from the left wall to the ceiling to the right wall. If I imagine the ceiling as a flat surface, the gecko's path would be a straight line from its position on the left wall to the fly's position on the right wall, passing through the ceiling. To calculate this, we can think of the ceiling as a rectangle and calculate the distance as follows:

The gecko is 11 feet from the front (x=11), 7 feet from the floor (z=7). The fly is 1 foot from the front (x=1), 1 foot from the floor (z=1). If the gecko crosses the ceiling, which is at z=8, then the vertical distance from the gecko to the ceiling is 8 - 7 = 1 foot, and from the ceiling to the fly is 8 - 1 = 7 feet. But this is in 3D, and we need to convert it to 2D.

Wait, if we unfold the left wall and the ceiling into a plane, then the gecko's position is at (x=11, z=7) and the fly's position, when projected onto the ceiling, would be at (x=1, z=8) because it's on the ceiling? No, the fly is on the right wall, not the ceiling. Wait, if we go from the left wall to the ceiling to the right wall, then on the ceiling, the gecko would reach a point that is on the way to the right wall. Let me think of it as two separate steps: from left wall to ceiling, and then from ceiling to right wall.

Let me calculate the distance from the gecko to a point on the ceiling, and from there to the fly. Let's call the point where the path meets the ceiling as (x, 8) on the ceiling. Then the total distance is the distance from (11, 0, 7) to (x, 8) on the ceiling, plus the distance from (x, 8) to (1, 10, 1) on the right wall.

But this is a minimization problem, where we need to find x that minimizes the total distance. However, this might be complex. Alternatively, since the shortest path should be a straight line in the unfolded net, we can find x such that the path is straight.

If we unfold the left wall and the ceiling into a single plane, then the gecko's position is (11, 7) on the left wall, and the fly's position on the right wall would be at (1, 10, 1). But how to represent this in the unfolded net?

Wait, if we unfold the left wall and the ceiling, the ceiling is now a horizontal surface. The right wall is at y=10, so if we imagine the ceiling extending in the y-direction, then the fly's position on the right wall is at (x=1, y=10, z=1). But on the unfolded net, this would be at (x=1, y=10) on the ceiling? Wait, no, because the fly is on the right wall, not the ceiling.

This is getting too tangled. Maybe I should look for an example of similar problems.

In many shortest path problems on a rectangular prism, the shortest path can be found by unfolding the prism into a flat net and measuring the straight line distance. For example, if the gecko needs to go from one point to another across two adjacent faces, the distance is the hypotenuse of a right triangle with sides equal to the sum of certain dimensions and the difference of others.

Let me try to apply this. Suppose the gecko goes from the left wall to the ceiling to the right wall. The total distance would be the hypotenuse of a right triangle where one leg is the distance along the length of the room (from x=11 to x=1, which is 10 feet) and the other leg is the distance from the gecko's height to the fly's height plus the height of the ceiling? Or something like that.

Wait, if we consider the gecko going from (11, 0, 7) to the ceiling, then to (1, 10, 1), the total horizontal distance would be from x=11 to x=1, which is 10 feet. The vertical distance would be from z=7 to z=8 (1 foot up) and then from z=8 to z=1 (7 feet down), totaling 8 feet. So the total distance would be sqrt(10² + 8²) = sqrt(164) ≈ 12.8 feet. But this is just a guess.

Alternatively, if the unwrapped net is such that the total horizontal distance is 10 feet and the total vertical distance is 7 (from z=7 to z=1) plus 1 (from z=7 to ceiling)? No, I'm not sure.

Wait, let's think of it this way: if the gecko crosses the ceiling, the path would form a rectangle when unfolded. The left wall and the ceiling form a rectangle of length 12 (x) and height 8 (z) + 10 (y)? No, it's more like the left wall (12 x 8) and the ceiling (12 x 10) combined. If we unfold them, the total height would be 8 + 10 = 18, and the length is 12. But the gecko's position is at (11, 7) on the left wall, and the fly is at (1, 10, 1) on the right wall. Wait, but where is the fly in this unfolded net?

If we imagine the left wall and the ceiling as part of the net, the right wall is not included. So perhaps we need to include it as well. If we unfold the left wall, ceiling, and right wall, the total height is 8 (left wall) + 10 (ceiling) + 8 (right wall) = 26? No, this is incorrect.

Alternatively, if the gecko goes from the left wall to the ceiling to the right wall, the total distance would be the hypotenuse of a right triangle with one leg being the length from x=11 to x=1 (10 feet) and the other leg being the height difference from z=7 to z=1 (6 feet) plus the height of the ceiling (8 feet)? No, this is not coherent.

Wait, perhaps the correct way is to think of the path as going from (11, 0, 7) to (1, 10, 1) through the ceiling. If we unfold the left wall and the ceiling into a plane, then the coordinates on this plane would be such that the ceiling extends in one direction. Let me assign coordinates: let's say the left wall is in the x-z plane, and the ceiling is in the x-y plane. If we unfold them such that the x-axis is common, then the total distance would be calculated by combining the z and y dimensions.

So, the gecko's position is at (11, 7) in x-z, and the fly's position is at (1, 10) in x-y, but they are on different walls. Wait, no, the fly is on the right wall, which is at y=10. If the ceiling is unfolded, then the right wall is at y=10, z=8? Not sure.

Alternatively, think of the total distance as a rectangle where the gecko travels from (11, 7) to (1, 10) in a combined x-y and x-z coordinate system. This is too vague.

At this point, I think I need to refer to the standard method for solving such problems, which is to consider the room as a box and find the shortest path by unfolding it into a plane.

For example, the shortest path from one point to another on a box can be found by considering different possible nets and calculating the distance for each.

Let me try to consider the path going across the left wall and the ceiling. To do this, the gecko would travel from the left wall to the ceiling, and then to the right wall. When you unfold the left wall and the ceiling into a single plane, the fly's position on the right wall is now on the ceiling. Wait, how?

If we unfold the left wall and the ceiling, then the right wall is adjacent to the ceiling. So, the fly's position on the right wall is at (1, 10, 1), but in this unfolded net, it would be at (1, 10) on the right wall, which is connected to the ceiling. If we represent this in the unfolded net, the total distance would be the straight line from (11, 7) on the left wall to (1, 10) on the right wall, with the ceiling in between.

But how to calculate this? If we imagine the left wall and ceiling as a single plane, the total height would be the height of the left wall (8) plus the width of the ceiling (10)? No, if the ceiling is attached to the top of the left wall, then the total height is 8 + 10 = 18, but the x-axis is still 12.

Wait, let me define the unfolded net as having the left wall (12x8) and the ceiling (12x10) attached to it. So the combined surface is a rectangle of 12x18. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10 + 8 = 18)? Wait, no.

Actually, if the ceiling is attached to the top of the left wall, then the ceiling's y-coordinate (which is 10) would be added to the left wall's z-coordinate (8). So, the fly's position on the right wall is at x=1, y=10, z=1. But in the unfolded net, the right wall is at y=10, so if we consider the ceiling to be in the y-direction, then the fly's position is at y=10, x=1, and z=1? This is confusing.

Maybe it's better to use the following method: when you unfold the left wall and the ceiling, the total vertical dimension is the height of the left wall (8) plus the width of the ceiling (10). The gecko's position is at (x=11, z=7) on the left wall, and the fly's position is at (x=1, y=10, z=1) on the right wall. To represent the fly's position in this unfolded net, we need to map it to the combined dimensions.

If we consider the left wall's z-axis and the ceiling's y-axis as a single dimension, then the fly's position would be at a distance of 8 (height of left wall) + 10 (width of ceiling) - 1 (z=1)? No, this is not linear.

Let me try to think of it as a 2D coordinate system where one axis is the length of the room (x from 0 to 12) and the other axis is a combined dimension of height (z) and width (y). If the gecko goes from the left wall to the ceiling, then the combined dimension would be z (from 0 to 8) + y (from 0 to 10). So, the gecko's position is at (x=11, z=7), and the fly's position is at (x=1, y=10, z=1). In this combined system, the z and y are combined into a single dimension. But how?

If we represent the combined dimension as 8 + 10 = 18, then the gecko's position is at (11, 7) and the fly's position is at (1, 10 + 1 = 11)? Wait, why 10 + 1? If the fly is on the right wall at z=1, then in the combined dimension, it would be 8 (height of left wall) + 1 (z=1 from the floor)? No, this is not correct.

I think I'm stuck here. Let me try to look for another approach.

Let me recall that in similar problems, the shortest path can be found by considering the room as a box and "flattening" it into a net. For example, if the gecko goes from the left wall to the ceiling to the right wall, the distance can be calculated as follows:

The gecko travels across the left wall and the ceiling. The total distance would be the hypotenuse of a right triangle where one leg is the length of the room minus the distances from the start and end points, and the other leg is the sum of the heights.

Wait, let me try this. The gecko is 1 foot from the back wall (x=11), and the fly is 1 foot from the front wall (x=1). The length between them is 11 - 1 = 10 feet. The gecko is 1 foot from the ceiling (z=7), and the fly is 1 foot from the floor (z=1). If the path goes across the ceiling, the vertical distance would be from z=7 to z=8 (1 foot) and from z=8 to z=1 (7 feet), totaling 8 feet. So the distance would be sqrt(10² + 8²) = sqrt(164) = 2√41 ≈ 12.806 feet.

But is this correct? Let me check.

If the gecko goes from (11, 0, 7) to the ceiling at (11, 0, 8), which is 1 foot, and then from there to (1, 10, 1), which would be a distance of sqrt((11-1)^2 + (0-10)^2 + (8-1)^2) = sqrt(100 + 100 + 49) = sqrt(249) ≈ 15.78 feet, which is longer. So this approach is incorrect.

Wait, but if we unfold the left wall and the ceiling, then the path is a straight line. So let me think of it as a 2D problem where the left wall and ceiling are flat. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10) on the right wall. But how?

If we imagine the left wall and ceiling as a single plane, then the right wall is at a distance of 10 feet (width of the room) from the left wall. Wait, no. The width of the room is 10 feet, but the gecko is on the left wall (y=0), and the fly is on the right wall (y=10). So the distance between the left and right walls is 10 feet.

If we unfold the left wall and the ceiling, then the gecko's position is at (x=11, z=7), and the fly's position is at (x=1, y=10, z=1). But if the ceiling is unfolded, the y-coordinate is now part of the vertical dimension. So the total vertical distance is the z-coordinate of the gecko (7) plus the y-coordinate of the fly (10) minus the z-coordinate of the fly (1)? No, this is not clear.

At this point, I think I need to accept that I'm struggling with visualizing the net and look for a different strategy.

Let me recall that when dealing with shortest paths on a cube, the shortest path can often be found by considering different possible unfoldings and calculating the Euclidean distance.

For this problem, the room is a rectangular prism, not a cube, but similar principles apply.

Let me consider the following: the gecko is on one side wall, and the fly is on the opposite side wall. To get from one to the other, the gecko can go across the ceiling, floor, or a combination of other walls.

Let me calculate the distance for the path that goes from the left wall across the ceiling to the right wall, and similarly for the floor.

For the ceiling path: If we imagine unfolding the left wall and the ceiling into a single plane, the gecko's position is at (11, 7) and the fly's position is at (1, 10). Wait, how?

Let me think of the combined height as the height of the left wall (8) + width of the ceiling (10) = 18. But the gecko is at z=7 (7 feet from the floor) on the left wall, and the fly is at 1 foot from the floor on the right wall, which is 1 foot from the floor. If we imagine the ceiling as extending in the y-direction, then the fly's position is at y=10 (width of the room) and z=1 (height from floor). But in the unfolded net, this would be at a distance of 8 (height of left wall) + 1 (height from floor on the right wall) = 9? No, this is not correct.

Wait, if we unfold the left wall and the ceiling, the height from the floor on the left wall is z=7, and on the ceiling, it would be y=10 (width) and z=8 (height). But the fly is on the right wall at z=1, which is 1 foot from the floor. If we imagine the right wall as being attached to the ceiling, then the fly's position would be at y=10 (width) and z=1 (height from floor), which in the unfolded net would be at y=10 and z=1, but how does that relate to the left wall's z=7?

Maybe I should think of it as follows: when you unfold the left wall and ceiling, the total vertical dimension is 8 (left wall height) + 10 (ceiling width) = 18. The gecko is 7 feet up on the left wall, and the fly is 1 foot up on the right wall. If the right wall is at the end of the ceiling, then the fly's position is at 10 feet in the width direction (y=10) and 1 foot in the height (z=1). But how do these combine?

If we consider the combined vertical dimension as z (from 0 to 8) and y (from 0 to 10), then the gecko's position is at (11, 7) and the fly's position is at (1, 10, 1). Wait, this is 3D, but if we flatten it into 2D, it would be a rectangle of (x, z + y)? No, this is too abstract.

Let me try to use the formula for the shortest path on a rectangular prism. The shortest path can be calculated by finding the minimal distance across different nets.

For example, if the gecko travels across the left wall and the ceiling, the distance can be calculated as follows:

The path would go from (11, 0, 7) to (1, 10, 1) via the ceiling. If we unfold the left wall and the ceiling, the total distance is the hypotenuse of a right triangle with one leg being the length between x=11 and x=1 (10 feet), and the other leg being the vertical distance from z=7 to z=8 (1 foot) plus the vertical distance from z=8 to z=1 (7 feet), totaling 8 feet. So the distance is sqrt(10² + 8²) = sqrt(164) = 2√41 ≈ 12.806.

Alternatively, if the gecko goes across the left wall and the floor, the vertical distance would be from z=7 to z=0 (7 feet) and from z=0 to z=1 (1 foot), totaling 8 feet. So the distance is also sqrt(10² + 8²) = sqrt(164). So both paths give the same distance.

But is there a shorter path through other faces?

Let me consider going from the left wall to the front wall to the right wall.

The front wall is at x=0. The gecko is at x=11 on the left wall, so the distance along the x-axis from the gecko to the front wall is 11 feet. The front wall is 10 feet wide (y) and 8 feet tall (z). The fly is at x=1, y=10, z=1. If the gecko goes from the left wall to the front wall to the right wall, the total distance would be the hypotenuse of a right triangle with one leg being the distance along the x-axis from 11 to 0 (11 feet) and the other leg being the distance from the gecko's position to the fly's position in y and z. But this is getting complex.

Let me try to calculate it. If the gecko goes from (11, 0, 7) to the front wall, then to (1, 10, 1). If we unfold the left wall and the front wall, the total distance would be the hypotenuse of a right triangle with one leg being the distance from x=11 to x=1 (10 feet) and the other leg being the distance from z=7 to z=1 (6 feet) plus the width of the front wall? No, this is not clear.

Alternatively, if the gecko goes from the left wall to the front wall (11 feet in x-direction), and then from the front wall to the right wall (1 foot in x-direction), but this is not helpful.

Alternatively, using the same method as before, if the gecko goes from left wall to front wall to right wall, the total distance would be the hypotenuse of a right triangle with one leg being the length from x=11 to x=1 (10 feet) and the other leg being the sum of the height from z=7 to z=1 (6 feet) and the width from y=0 to y=10 (10 feet)? No, this is not correct.

At this point, I think the shortest path is likely the one that goes across the ceiling or floor, giving a distance of sqrt(10² + 8²) = sqrt(164) = 2√41. But I need to confirm.

Wait, let's calculate the distance if the gecko goes from the left wall to the ceiling to the right wall. If we consider the left wall and the ceiling as one plane, and the right wall as another, the distance would be the straight line from (11, 7) to (1, 10 + 8 - 1)? No, this is not coherent.

Wait, let me try to use the method of reflecting the fly's position across the ceiling. If the gecko goes from the left wall to the ceiling to the right wall, we can reflect the fly's position across the ceiling to find the straight line.

If the fly is at (1, 10, 1), its reflection across the ceiling (z=8) would be at (1, 10, 15). The distance from the gecko's position (11, 0, 7) to the reflected point (1, 10, 15) would be sqrt((11 - 1)^2 + (0 - 10)^2 + (7 - 15)^2) = sqrt(100 + 100 + 64) = sqrt(264) = 2√66 ≈ 16.248, which is longer than the previous calculation, so this is not helpful.

Alternatively, if we reflect the gecko's position across the ceiling, but this is getting too complex.

Wait, let me think of the problem as a 2D problem by considering the room's dimensions.

If the gecko goes from the left wall to the ceiling to the right wall, we can imagine flattening the left wall and ceiling into a plane. The total distance would be the hypotenuse of a right triangle with one leg being the distance between the x-coordinates (11 to 1 = 10 feet) and the other leg being the distance in the y-z plane. The y-z plane here would be the height of the gecko (7 feet) plus the height from the ceiling to the fly (1 foot), but this is not correct.

Wait, if the gecko is at (11, 0, 7) and the fly is at (1, 10, 1), and the path goes across the ceiling, then on the unfolded net, the distance would be from (11, 7) to (1, 10 + 8 - 1)? No, I'm not sure.

Let me try to think of the unfolded net as having the left wall and the ceiling combined. The left wall has dimensions 12 x 8, and the ceiling has 12 x 10. If we lay them flat, the total height would be 8 + 10 = 18, and the length is 12. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10) on the right wall. Wait, if the right wall is at the end of the ceiling, then the fly's position is at (1, 10) on the right wall, which is 10 feet in the y-direction (width) and 1 foot in the z-direction (height). But in the unfolded net, the total height is 18, so the fly's position would be at (1, 10 + 8 - 1) = (1, 17)? No, this is not logical.

I think I need to accept that I'm not able to visualize the correct unfolding and look for another way.

Let me consider that the shortest path will involve going across two adjacent walls. Let's calculate the distance for each possible pair of walls and choose the smallest.

Possible pairs:

1. Left wall and ceiling: distance = sqrt( (11 - 1)^2 + (7 + 1)^2 ) = sqrt(100 + 64) = sqrt(164)

Wait, how did I get (7 + 1)? If the gecko goes from z=7 on the left wall to z=8 on the ceiling (1 foot), and then from z=8 to z=1 on the right wall (7 feet), but in the unfolded net, it's a straight line. If the total vertical distance is 7 + 1 = 8, then the distance is sqrt(10² + 8²) = sqrt(164).

2. Left wall and floor: similar calculation, sqrt(10² + (7 + 1)^2) = sqrt(164).

3. Left wall and front wall: if the gecko goes from left wall to front wall to right wall. The front wall is at x=0, so the distance from x=11 to x=0 is 11 feet. The vertical distance would be from z=7 to z=1, which is 6 feet, and the width from y=0 to y=10 is 10 feet. But this would form a right triangle with legs 11 and 10 + 6 = 16? No, this is not correct.

Wait, if the gecko goes from left wall to front wall to right wall, it would traverse the left wall to front wall (which is 11 feet in x), and then front wall to right wall (1 foot in x). But the vertical distance would be from z=7 to z=1, which is 6 feet. So the total distance would be sqrt(11² + 6²) + sqrt(1² + 10²)? No, this is not correct.

Alternatively, if we unfold the left wall and front wall, the distance would be sqrt((11 + 1)^2 + (7 + 1)^2) = sqrt(144 + 64) = sqrt(208) = 4√13 ≈ 14.422, which is longer than sqrt(164) ≈ 12.806.

4. Left wall and back wall: the gecko is already one foot from the back wall (x=11), so going to the back wall would be 0 feet, but the fly is on the opposite wall, so this might not help.

5. Left wall and right wall: but they are opposite, so the gecko cannot go directly.

So, the shortest distance seems to be sqrt(164), which is 2√41. Let me check if there are other possible paths.

Wait, what if the gecko goes across the ceiling and the front wall? Let me try that.

If the gecko goes from left wall to ceiling to front wall to right wall, but this would be a longer path. Let me calculate it.

But this is getting too complex. Based on the previous calculations, the shortest path seems to be sqrt(164). Let me check if this is correct.

Wait, let me verify with another approach. Let's consider the room as a box with dimensions 12x10x8. The gecko is at (11, 0, 7) and the fly is at (1, 10, 1). The shortest path on the surfaces can be found by considering the net where the path crosses two adjacent faces.

If we consider the left wall (12x8) and the ceiling (12x10) as a single net, the distance would be the hypotenuse of a right triangle with one leg being the distance between x=11 and x=1 (10 feet) and the other leg being the distance between z=7 and z=1 (6 feet) plus the width of the ceiling (10 feet)? No, this is not correct.

Wait, another way: if we unfold the left wall and the ceiling such that the gecko's path is a straight line, the total vertical distance would be the distance from z=7 on the left wall to z=1 on the right wall, which is 7 - 1 = 6 feet, but since it goes through the ceiling, it would be 8 - 7 + 1 = 2 feet? No, this is confused.

At this point, I think I need to proceed with the sqrt(164) answer, which is 2√41, and see if it matches with the correct answer.

But let me check another possibility: if the gecko goes across the ceiling and the right wall, but I think it's the same as going across the ceiling.

Alternatively, let me calculate the distance if the gecko goes across the left wall and the ceiling, and the right wall is part of the ceiling. Wait, the right wall is at y=10, so if we imagine unfolding the ceiling and the right wall, the total width would be 10 (ceiling) + 8 (right wall) = 18. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 1) on the right wall. Wait, no.

If we unfold the ceiling and the right wall, the gecko's position on the left wall is now at (11, 7) on the left wall, and the fly is at (1, 1) on the right wall. The distance would be sqrt((11 - 1)^2 + (7 + 10 + 1)^2)? No, this is not correct.

I think I need to accept that the shortest path is 2√41, which is sqrt(164), and check if this is the correct answer.

But let me think again: if the gecko goes from (11, 0, 7) to (1, 10, 1) via the ceiling, the path would be a straight line in the unfolded net. If we consider the left wall and ceiling as a single plane, then the total vertical dimension is 8 (height of left wall) + 10 (width of ceiling) = 18. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10) on the right wall. Wait, if the right wall is at the end of the ceiling, then the fly's position is at x=1, y=10, which in the unfolded net would be at x=1, y=10 + 8 = 18? No.

Alternatively, if the fly is on the right wall at (1, 10, 1), and we're in the unfolded net of left wall and ceiling, then the fly's position is at (1, 10) on the right wall, which is at a distance of 10 feet in y and 1 foot in z. But how does this translate to the unfolded net?

I think I'm going in circles. Given that I've spent a lot of time and the most plausible answer seems to be sqrt(164) or 2√41, I'll go with that, but I need to check if there's a shorter path.

Wait, let me try another unfolding. Suppose the gecko goes from the left wall to the front wall to the right wall. If we unfold the left wall and the front wall, the total distance would be the hypotenuse of a right triangle with one leg being the distance from x=11 to x=1 (10 feet) and the other leg being the distance from z=7 to z=1 (6 feet) plus the width of the front wall (10 feet)? No, this is not correct.

Wait, if we unfold the left wall and the front wall into a single plane, the total dimensions would be 12 (length) by 8 (height) + 10 (width) = 18. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10) on the right wall. The distance would be sqrt((11 - 1)^2 + (7 + 10 - 1)^2) = sqrt(100 + 16²) = sqrt(100 + 256) = sqrt(356) ≈ 18.87, which is longer.

Therefore, the shortest path is indeed sqrt(164) = 2√41.

But let me verify with another possible unfolding. What if the gecko goes across the ceiling and the right wall? Let me imagine unfolding the ceiling and the right wall. The total dimensions would be 12 (length) by 8 (height of right wall) + 10 (width of ceiling) = 18. The gecko's position is at (11, 7) on the left wall, which is now at (11, 8 + 7 - 8 = 7)? No, this is not clear.

Alternatively, if the gecko goes from the left wall to the ceiling to the right wall, the unfolded net is left wall (12x8) + ceiling (12x10) + right wall (12x8). If we lay them out in a straight line, the total height is 8 + 10 + 8 = 26. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 1) on the right wall. The distance would be sqrt((11 - 1)^2 + (7 + 10 + 1)^2) = sqrt(100 + 18²) = sqrt(100 + 324) = sqrt(424) ≈ 20.59, which is longer.

Thus, the shortest path is indeed sqrt(164), which simplifies to 2√41.

Wait, but let me check if there's a different unfolding where the distance is smaller. For example, what if the gecko goes across the ceiling and the back wall? Or other combinations?

Let me try another possibility: the gecko goes from the left wall to the back wall to the right wall. The back wall is at x=12. The gecko is at x=11, so it's 1 foot from the back wall. If it goes to the back wall, then to the right wall. The distance would be sqrt(1² + (7 + 1)^2) + sqrt(12 - 1)^2 + ... No, this is not helpful.

Alternatively, if we unfold the back wall and the right wall, but this is getting too complex.

Given all the above, I think the shortest path is 2√41, which is sqrt(164). Let me check if this is correct.

Wait, but let me calculate 10² + 8² = 100 + 64 = 164. Yes, sqrt(164) is approximately 12.806.

But I recall that in similar problems, sometimes the shortest path involves going across three faces, but in this case, it seems that going across two faces is sufficient.

Wait, but let me think: if the gecko goes from the left wall to the ceiling to the right wall, the actual distance is sqrt((11-1)^2 + (7 + 10 - 1)^2)? No, why 7 + 10 - 1?

Wait, if the gecko is at (11, 0, 7) on the left wall, and the fly is at (1, 10, 1) on the right wall, and the path goes across the ceiling. If we unfold the left wall and the ceiling, then the total vertical distance is from z=7 on the left wall to z=1 on the right wall, passing through the ceiling. If we consider the ceiling as extending in the y-direction, the vertical distance would be 7 (from z=7 to z=8) + 1 (from z=8 to z=1) = 8. So the distance is sqrt(10² + 8²) = sqrt(164). This seems to be the correct calculation.

Yes, this makes sense. So the shortest path is 2√41, which is the simplified radical form.

**Final Answer**
The length of the shortest path is \boxed{2\sqrt{41}}.

To find the shortest path a gecko can take from one point to another on the **surface** of a rectangular room, it's helpful to **unfold** the room into a 2D net. This allows us to calculate the straight-line distance (which represents the path along the surfaces of the room) using the **Pythagorean Theorem**.

---

### **1. Understanding the Room and Coordinates**

We are given:
- **Room dimensions**: 12 feet (length × x-axis), 10 feet (width × y-axis), and 8 feet (height × z-axis).
- **Gecko's starting position**: On the **left wall** (y = 0), one foot from the ceiling (z = 7) and one foot from the **back wall** (x = 11). So,
  - Start: (11, 0, 7)
- **Fly's position**: On the **opposite right wall** (y = 10), one foot from the **floor** (z = 1) and one foot from the **front wall** (x = 1). So,
  - End: (1, 10, 1)

---

### **2. Choosing the Shortest Path**

Since the gecko is on the **left wall** and the fly is on the **opposite right wall**, the shortest path likely involves **moving across two adjacent faces**—for example, the **left wall** and the **ceiling**. This way, the gecko travels across the **least amount of surface area** and avoids unnecessary detours.

If the gecko crosses the ceiling, the path becomes a straight line in a 2D unfolded net, where:
- The **horizontal distance** between the gecko and the fly is from x = 11 to x = 1:
  $$
  \text{Horizontal distance} = 11 - 1 = 10 \text{ feet}
  $$
- The **vertical distance** is from z = 7 (on the left wall) to z = 1 (on the right wall), passing through the ceiling. The total vertical distance is:
  $$
  \text{Vertical distance} = (8 - 7) + (8 - 1) = 1 + 7 = 8 \text{ feet}
  $$

Now, treat this as a right triangle with:
- Base: 10 feet
- Height: 8 feet

---

### **3. Computing the Shortest Path**

Using the **Pythagorean Theorem**:
$$
\text{Shortest path} = \sqrt{10^2 + 8^2} = \sqrt{100 + 64} = \sqrt{164}
$$

Simplify:
$$
\sqrt{164} = \sqrt{4 \times 41} = 2\sqrt{41}
$$

---

### **Final Result**

$$
\boxed{2\sqrt{41}}
$$

This is the **shortest possible path** the gecko can take on the surface of the room to reach the fly, assuming it cannot jump and is allowed to walk across the floor, ceiling, and walls.

[Assistant]
Explanation:
Okay, so there's this gecko in a room, and I need to figure out the shortest path it can take to get from one point to another without jumping. The room is 12 feet long, 10 feet wide, and 8 feet tall. Let me try to visualize this. The gecko is on a side wall, which is 10 feet by 8 feet. Wait, the room has length, width, and height. Let me confirm: usually, in a room, length, width, and height. But the problem mentions a side wall that's 10 by 8. Wait, the room is 12 feet long, 10 feet wide, and 8 feet tall. So the walls would be:

- The front and back walls: length x height = 12 ft by 8 ft
- The left and right walls: width x height = 10 ft by 8 ft
- The floor and ceiling: length x width = 12 ft by 10 ft

So the gecko is on a side wall (which is width x height, so 10 by 8). The gecko is one foot from the ceiling and one foot from the back wall. Wait, the back wall is 12 by 8? Or is the back wall part of the length? Let me think again.

The problem says: the gecko is on a side wall (10' by 8'), one foot from the ceiling and one foot from the back wall (12' by 8'). So the side wall is 10 by 8, meaning that it's either the left or right wall. The back wall is 12 by 8, which would be the back wall (assuming that the front wall is also 12 by 8, and the left and right are 10 by 8). So the gecko is on a side wall (left or right, 10 by 8), one foot from the ceiling (so 8 - 1 = 7 feet from the floor?) and one foot from the back wall. Wait, the back wall is 12 by 8, so if the gecko is one foot from the back wall, that would be along the length of the room? Let me try to sketch mentally.

Let me consider the room as a rectangular box with length 12, width 10, height 8. Let me assign coordinates. Let me think of the room with coordinates (x, y, z), where x is along the length (from front to back), y is along the width (left to right), and z is height (floor to ceiling). So front wall is at x=0, back wall is at x=12, left wall is at y=0, right wall is at y=10, floor at z=0, ceiling at z=8.

So the gecko is on a side wall (which is y=0 or y=10). The problem says "one foot from the ceiling and one foot from the back wall". So if the gecko is on a side wall, which is y=0 or y=10. Let me assume it's on the left side wall (y=0). Then "one foot from the back wall" would mean along the x-axis. Since back wall is at x=12, so one foot from back wall would be at x=12 - 1 = 11 feet. Also, "one foot from the ceiling" would be z=8 - 1 = 7 feet. So the gecko is at position (11, 0, 7). Wait, but is that correct?

Wait, if the gecko is on the left wall (y=0), then the coordinates would be (x, 0, z). The back wall is at x=12, so one foot from back wall is x=12 - 1 = 11. And one foot from ceiling is z=8 - 1 = 7. So yes, (11, 0, 7).

Then the fly is on the opposite side wall, which would be the right wall (y=10). The fly is one foot from the floor and one foot from the front wall. Front wall is at x=0, so one foot from front wall is x=0 + 1 = 1 foot. One foot from floor is z=0 + 1 = 1 foot. Therefore, the fly is at (1, 10, 1).

So now, the gecko is at (11, 0, 7), and the fly is at (1, 10, 1). The gecko can walk on the ceiling and walls, but not jump. So the question is to find the shortest path from (11, 0, 7) to (1, 10, 1) along the surfaces of the room.

This seems like a classic shortest path on a surface problem, which is often solved by unfolding the room into a net, and then finding the straight line distance on the net.

So, to figure out the shortest path, we need to figure out which walls or ceiling the gecko will traverse. Since the gecko is on the left wall, and the fly is on the right wall, the shortest path might go across the ceiling or the floor or some combination of walls.

But since the gecko can walk on the ceiling and walls, perhaps the shortest path would involve unfolding the walls into a plane. Let me think about different possible unfoldings.

First, let me note that the gecko is on the left wall (y=0), and the fly is on the right wall (y=10). The straight-line distance through the room would be sqrt[(12 - 2)^2 + (10 - 0)^2 + (8 - 6)^2]? Wait, no, that's not correct because the gecko can't go through the room, it has to stay on the surfaces.

So, the key idea is to find a way to unfold the walls such that the path becomes a straight line. There are multiple ways to unfold the room into a plane, and we need to find the one that gives the shortest distance.

Let me try to think step by step.

First, the gecko is on the left wall (y=0). Let me imagine unfolding the walls adjacent to the left wall (the floor, ceiling, and front/back walls) into a plane. Similarly, for the right wall.

But since the gecko needs to reach the opposite side wall, maybe the path goes from the left wall to the ceiling or floor, then to the right wall? Or maybe across the front or back wall?

Alternatively, perhaps the gecko can go from the left wall, cross the ceiling to the right wall, or cross the floor to the right wall. Let me try to calculate both possibilities.

First, let's think about going from the left wall to the ceiling, then to the right wall. Alternatively, maybe going from the left wall to the floor then to the right wall.

Alternatively, the gecko could go from the left wall to the front wall, then to the right wall, or to the back wall?

But since the gecko is one foot away from the back wall (at x=11), and the fly is one foot away from the front wall (at x=1), maybe the path goes through the front wall?

But let me think. Let me try to visualize unfolding the walls.

Let me first think about the possibility of unfolding the left wall and the ceiling into a flat surface. So if we unfold the left wall (y=0) and the ceiling (z=8), how would that look? If we imagine the left wall (which is 10ft wide (y-axis) by 8ft tall (z-axis)) and the ceiling (which is 12ft long (x-axis) by 10ft wide (y-axis)). But how do they connect? The ceiling connects to the left wall along the top edge of the left wall (z=8, y=0). If we unfold the left wall and the ceiling, we can imagine rotating the ceiling so that it's adjacent to the left wall. But maybe there's a better way.

Alternatively, maybe if we unfold the left wall, the back wall, and the ceiling? But maybe I need to consider different combinations.

Alternatively, think of the room as a box, and the shortest path will cross two adjacent faces. For example, from the left wall to the ceiling, then to the right wall. Or from the left wall to the floor, then to the right wall. Or from the left wall to the front wall, then to the right wall, etc.

Let me try to consider different cases.

Case 1: Gecko goes from left wall to ceiling to right wall.

To calculate this, we need to imagine unfolding the left wall and the ceiling into a plane. So, if we take the left wall (y=0) and unfold the ceiling (z=8) adjacent to it. The left wall is 10ft (width) by 8ft (height). The ceiling is 12ft (length) by 10ft (width). If we unfold the ceiling upwards from the left wall, then the total surface would be a rectangle with length equal to the height of the left wall (8ft) plus the length of the ceiling? Wait, no, maybe.

Wait, when you unfold the left wall (y=0) and the ceiling (z=8), the left wall is vertical, and the ceiling is horizontal. If we unfold them into a plane, the left wall would be connected along the top edge to the ceiling. So if you imagine flattening the ceiling so that it becomes adjacent to the left wall, then the combined surface would be a rectangle. Let me think of coordinates. On the left wall, the gecko is at (x=11, y=0, z=7). Wait, but in terms of the left wall, which is y=0, x ranges from 0 to 12, z from 0 to 8.

But when we unfold the left wall and the ceiling, the ceiling is attached along the top edge (z=8) of the left wall. So if we unfold the ceiling by rotating it 90 degrees around the top edge, then the ceiling would be adjacent to the left wall. So in this case, the left wall and the ceiling form a larger rectangle. The left wall has dimensions 12 (x) by 8 (z). The ceiling has dimensions 12 (x) by 10 (y). But when unfolded, how do they connect? Wait, maybe I need to think of the left wall's y-axis? Wait, no, the left wall is at y=0, so its coordinates are x and z. The ceiling is at z=8, and its coordinates are x and y. So when you unfold them, you need to align the edge where they meet.

The edge where the left wall meets the ceiling is at y=0, z=8. So if you unfold the ceiling so that it's adjacent to the left wall, you can imagine the ceiling extending from the left wall. However, since the ceiling is 10 feet wide (y-direction) and 12 feet long (x-direction), but the left wall is 12 feet long (x-direction) and 8 feet tall (z-direction). So if you unfold the ceiling so that it's adjacent to the left wall along the top edge (z=8, y=0), then the ceiling would be folded outwards from the left wall. Therefore, the combined surface would be a rectangle that is 12 feet (x) in length, and 8 (z) + 10 (y)? Wait, but that might not be right. Wait, when you unfold the ceiling, you can think of it as extending in the y-direction from the left wall's edge. Wait, maybe it's better to think of the left wall and the ceiling as forming a larger plane.

Alternatively, perhaps if you unfold the left wall and the ceiling into a plane, the total distance would be calculated by considering the gecko moving from its position on the left wall to some point on the ceiling, then to the fly's position on the right wall. But this seems complex. Maybe it's better to use the net method.

Alternatively, think of the path as crossing two adjacent faces: the left wall and the ceiling. If I can represent those two faces as a single plane, then the straight-line distance would be the shortest path.

Similarly, if the gecko crosses the left wall and the front wall, or left wall and the floor, etc.

Alternatively, think of unfolding the room in such a way that the gecko's path becomes a straight line. Since the gecko is on the left wall, and the fly is on the right wall, maybe the path crosses the ceiling or the floor.

Let me consider unfolding the left wall and the ceiling. Let me imagine the left wall (size 12x8) and the ceiling (12x10). When unfolded, the left wall and ceiling are adjacent. But how?

Wait, the left wall is at y=0, with x from 0 to 12, z from 0 to 8. The ceiling is at z=8, with x from 0 to 12, y from 0 to 10. So if you unfold the ceiling so that it's attached to the left wall's top edge (z=8), then the ceiling would be attached along the top of the left wall. Therefore, if you were to lay the ceiling flat next to the left wall, the combined surface would be a rectangle with dimensions x from 0 to 12, and z from 0 to 8 (left wall) plus y from 0 to 10 (ceiling)? Wait, maybe not. Wait, if we consider the left wall as a vertical rectangle and the ceiling as a horizontal rectangle attached to its top, then in the unfolded net, the vertical left wall and the horizontal ceiling would form a larger rectangle with dimensions length in x-direction 12, and height in z and y directions? Wait, this is getting confusing.

Maybe it's easier to think of coordinates. Suppose we unfold the left wall and the ceiling into a plane. Let me assign coordinates to each face.

Original coordinates:

Left wall (y=0): x (from 0 to 12), z (from 0 to 8).

Ceiling (z=8): x (from 0 to 12), y (from 0 to 10).

When we unfold the ceiling so that it's adjacent to the left wall, the ceiling is now in the same plane as the left wall. Let me imagine "flattening" the ceiling so that it's next to the left wall. So, if you take the ceiling and fold it down onto the left wall, but actually, to make it a plane, it's like taping the ceiling to the left wall along the top edge. However, in terms of coordinates, the ceiling's y-coordinate is perpendicular to the left wall's z-coordinate. To flatten them, we need to represent them as a single plane.

Alternatively, think of the left wall as having coordinates (x, z) where 0 ≤ x ≤ 12, 0 ≤ z ≤ 8. When unfolded with the ceiling, which has coordinates (x, y) where 0 ≤ x ≤ 12, 0 ≤ y ≤ 10. To represent them on the same plane, we can imagine that the ceiling is now in the same plane as the left wall, but with y and z coordinates combined. Wait, maybe not. Alternatively, we can consider reflecting coordinates.

Alternatively, think of the ceiling as extending from the left wall's top edge. So, if we consider moving from the left wall to the ceiling, the path would cross from (x, z) on the left wall to (x, y) on the ceiling, but since they are adjacent, we can represent this as a single plane where the ceiling is "folded" over the left wall.

But this is abstract. Maybe I can use the method of reflecting the room or considering the net as a combination of walls.

Wait, here's a better method: to find the shortest path on the surface, imagine unfolding the room so that the start and end points lie on the same plane, and then the straight line distance on that plane is the shortest path.

So, for example, if the gecko goes from left wall to ceiling to right wall, we can imagine unfolding the left wall and the ceiling into a single plane, and then the fly is on the right wall. But the right wall is not directly connected to the ceiling or left wall. Wait, the right wall is connected to the ceiling. So if we unfold the ceiling, then the right wall is adjacent to the ceiling.

Alternatively, perhaps the gecko goes from the left wall across the ceiling to the right wall. So, to represent this path as a straight line, we need to unfold the left wall and the ceiling and the right wall into a single plane. Wait, but how?

Alternatively, think of the left wall, ceiling, and right wall as a strip. If we unfold the left wall (y=0), the ceiling (z=8), and the right wall (y=10) all in one plane. But the ceiling is connected to both the left and right walls. If we lay the left wall, ceiling, and right wall flat, they would form a larger rectangle where the left wall is connected to the ceiling, and the ceiling is connected to the right wall. Therefore, the total length in the x-direction would be the same, but the height would be the height of the wall (8ft) plus the width of the ceiling (10ft)? Wait, no.

Wait, if you unfold the left wall, ceiling, and right wall, you can imagine the left wall (12x8), then the ceiling (12x10), then the right wall (12x8), connected along the x-axis. Wait, but the left wall and right wall are both connected to the ceiling. So when unfolded, they would form a sort of L-shape? No, if you lay them flat, they would be a straight line? Wait, no. If you take the left wall and unfold the ceiling so that it's adjacent to the left wall, then unfold the right wall adjacent to the ceiling. So, the left wall, ceiling, right wall would form a sort of long rectangle?

Wait, let me think of the left wall as a vertical rectangle, then the ceiling is horizontal, attached to the top of the left wall, and then the right wall is vertical, attached to the end of the ceiling. But actually, the right wall is also vertical, but connected to the ceiling. If you unfold the ceiling between the left and right walls, then the left wall, ceiling, right wall would form a flat plane where the total dimensions are?

In terms of length along x-axis: left wall is 12 feet (x from 0 to 12), the ceiling is 12 feet (x from 0 to 12), and the right wall is 12 feet (x from 0 to 12). But if we combine them, it's more about the other dimensions.

Alternatively, think of the left wall as having width in the z-direction (height) and the ceiling as having width in the y-direction. If we unfold them, we can represent them as a single plane with coordinates combining z and y. Wait, this is getting too abstract. Maybe it's better to use coordinates.

Suppose we imagine the gecko's starting point (11, 0, 7) on the left wall. If we unfold the left wall and the ceiling into a plane, then on this plane, the coordinates of the gecko would be (x, z) where z is height, and then the ceiling is unfolded so that it's adjacent. If we want to go from the left wall to the ceiling to the right wall, we need to consider coordinates on both.

Alternatively, another approach is to consider that when moving across two adjacent faces, the path will be a straight line in the unfolded net. Therefore, to calculate the distance, we can calculate the straight line between the start and end points on the net.

But to do that, we need to figure out which faces are involved in the shortest path. Since the gecko is on the left wall and the fly is on the right wall, the shortest path may go across the ceiling or the floor, or across the front/back wall.

Let me try considering different possibilities:

Possibility 1: The gecko goes from left wall to ceiling to right wall.

Possibility 2: The gecko goes from left wall to floor to right wall.

Possibility 3: The gecko goes from left wall to front wall to right wall.

Possibility 4: The gecko goes from left wall to back wall to right wall.

Possibility 5: The gecko goes across multiple walls, like left wall to ceiling to front wall to right wall, etc. But since we want the shortest path, it's likely that it goes through two walls.

So let's evaluate Possibility 1 and 2 first.

Starting with Possibility 1: Left wall to ceiling to right wall.

If we unfold the left wall and the ceiling so that they are on the same plane, then the gecko's starting point is on the left wall, and the fly's position on the right wall is now on the ceiling's side? Wait, not sure.

Wait, to make this concrete, let's think of the left wall as one face and the ceiling as another face. Let me imagine unfolding the left wall (y=0) and the ceiling (z=8) into a single plane. If we do that, then the gecko's position on the left wall is still at (x=11, z=7). However, the fly is on the right wall (y=10), which is not directly connected to the ceiling unless we also unfold the right wall. Therefore, to have the fly on the same plane as the gecko's path, we need to unfold the ceiling and the right wall as well.

Alternatively, think about unfolding the left wall, ceiling, and right wall into a single plane. If we do that, then the left wall, ceiling, and right wall form a sort of 3D shape that can be flattened into a 2D plane. The left wall is connected to the ceiling, and the ceiling is connected to the right wall. When unfolded, this would create a flat surface where the left wall and right wall are on opposite ends, connected via the ceiling.

In such a case, the total width (in the direction perpendicular to the x-axis) would be the height of the left wall (8 feet) plus the width of the ceiling (10 feet)? Wait, no. Wait, if we lay the left wall, ceiling, and right wall flat, the dimensions would be?

Wait, the left wall has dimensions 12 feet (x-direction) by 8 feet (z-direction). The ceiling has 12 feet (x-direction) by 10 feet (y-direction). The right wall has 12 feet (x-direction) by 8 feet (z-direction). When unfolded, the ceiling is between the left and right walls? If we lay them out so that the left wall is adjacent to the ceiling along the top edge (z=8), and the ceiling is adjacent to the right wall along its side (y=10). Therefore, if you lay this all flat, the total height in the y-z direction would be the height of the left wall (8ft) plus the width of the ceiling (10ft)? Wait, not quite. Wait, when unfolded, the left wall and ceiling are in one plane, and the ceiling and right wall are in another? No, it's all in one plane. Hmm.

Alternatively, think of the left wall as a vertical rectangle, and then the ceiling is like a horizontal rectangle connected to the top of the left wall. Then the right wall is connected to the end of the ceiling. But in terms of coordinates, if we create a net where the left wall is on the left, the ceiling is above it, and the right wall is to the right of the ceiling, then the entire net would be a sort of T-shape? Wait, maybe not. Alternatively, imagine the left wall and ceiling forming a vertical rectangle with the ceiling folded out. If we then fold the right wall out from the ceiling, it becomes a different orientation.

This is getting too vague. Let me try using coordinates with unfolding.

Let me consider unfolding the left wall (y=0) with the ceiling (z=8). When unfolded, the ceiling is adjacent to the left wall. So, the left wall's top edge (z=8) is connected to the ceiling's bottom edge (z=8). Therefore, if we were to represent this as a single plane, we can think of the left wall as going from z=0 to z=8, and the ceiling extending in the y-direction from y=0 to y=10. Wait, but in terms of coordinates, if we unfold the ceiling so that it is in the same plane as the left wall, then the coordinates can be combined.

Alternatively, if we think of the left wall's z-coordinate as the vertical axis and the ceiling's y-coordinate as the horizontal axis. Wait, maybe not. Let me try to assign coordinates.

Let me denote that on the left wall (y=0), the gecko is at position (x=11, z=7). The fly is on the right wall (y=10) at (x=1, z=1). If we unfold the left wall and the ceiling, the ceiling is adjacent to the left wall. Let me imagine the ceiling being laid flat so that it's adjacent to the left wall's top edge. So the ceiling is now extending in the y-direction from the left wall. Therefore, the coordinates on the unfolded net would be:

- On the left wall: (x, z) where 0 ≤ x ≤ 12, 0 ≤ z ≤ 8.

- On the ceiling: (x, y) where 0 ≤ x ≤ 12, 0 ≤ y ≤ 10, but in the unfolded net, this is adjacent to the left wall's top edge. So if we consider the left wall as having z from 0 to 8, the ceiling would be in a different dimension. Maybe we can map the ceiling's y to z? Wait, if we unfold the ceiling down from the left wall, then the ceiling's y becomes an extension in the z-direction? Not sure. Maybe this is getting too confusing.

Alternatively, think of the path as going from the gecko's starting point on the left wall, across the ceiling, to the fly's position on the right wall. To compute the distance, imagine that the ceiling is a rectangle (12x10) and the left and right walls are connected to it.

But how to model this? Perhaps, if we consider the three-dimensional coordinates, but when unfolded, the ceiling is flat. Let me think of the gecko's path as moving from (11, 0, 7) on the left wall to some point on the ceiling, then to (1, 10, 1) on the right wall. However, to find the shortest path, it's equivalent to finding a straight line in the unfolded net.

Alternatively, think of the ceiling as a bridge between the left and right walls. If we unfold the left wall, ceiling, and right wall into a single plane, the total horizontal space (in the x-direction) is still 12 feet, but the vertical space would be the height of the left wall (8 ft) plus the width of the ceiling (10 ft)? Wait, no, the width of the ceiling is in the y-direction, which is perpendicular to the x-direction. This is getting too tangled.

Wait, here's a different approach. The problem is similar to finding the shortest path on the surface of a rectangular prism between two points. The standard method is to consider different possible unfoldings and calculate the distance for each, then take the minimum.

In order to do that, I need to figure out which faces the gecko would traverse. Since the gecko is on the left wall and the fly is on the right wall, the shortest path will go through two adjacent faces. The two faces could be:

1. Left wall and ceiling (path goes from left wall to ceiling to right wall)
2. Left wall and floor (path goes from left wall to floor to right wall)
3. Left wall and front wall (path goes from left wall to front wall to right wall)
4. Left wall and back wall (path goes from left wall to back wall to right wall)

Also, maybe the path goes through three faces, but probably not necessary because the shortest path would cross two faces.

Let me calculate the distances for each of these cases.

Case 1: Left wall to ceiling to right wall.

To model this, imagine unfolding the left wall and the ceiling into a single plane. Then the fly would be on the right wall, which is connected to the ceiling. Wait, so maybe we need to include the right wall as well?

Alternatively, think of the left wall, ceiling, and right wall as a single plane. When unfolded, the left wall is connected to the ceiling, and the ceiling is connected to the right wall. Therefore, the total dimensions would be the x-direction is still 12 feet (from front to back), but the other dimension is the height of the left wall (8 feet) plus the width of the ceiling (10 feet)? Wait, no. Wait, if we take the left wall (which is 12x8), the ceiling (12x10), and the right wall (12x8), when unfolded, they form a shape that's 12 feet in length, and the height is 8 + 10 = 18 feet? Wait, but the right wall is connected to the back of the ceiling. Wait, no, the ceiling is in the middle.

Wait, maybe if we unfold the left wall, the ceiling, and the right wall into a single plane, the total height would be the height of the left wall (8) plus the width of the ceiling (10) plus the height of the right wall (8)? No, because the right wall is connected to the ceiling's side. Let me try to visualize.

Suppose we have the left wall, then the ceiling, then the right wall. If you unfold them so that they are all in the same plane, the left wall is vertical, the ceiling is horizontal, and the right wall is vertical again. If you lay them out flat, the total "height" would be the height of the left wall (8) plus the width of the ceiling (10) plus the height of the right wall (8)? That seems too much. Wait, no. When you unfold them, the ceiling is between the left wall and the right wall. So, if you have the left wall, then the ceiling, then the right wall, they are connected along the x-axis. Therefore, the total "width" in the y-z plane would be?

This is really tricky without a diagram. Maybe I need to use a different method.

Let me recall that in such problems, the shortest path can be found by reflecting the room. For example, reflecting the fly's position across the walls and finding the straight line.

Alternatively, think about the net. For instance, if the gecko goes from the left wall to the ceiling to the right wall, then we can imagine folding the left wall and the ceiling into a single plane, then the fly's position on the right wall is... Hmm. Wait, let me consider coordinates again.

Let me think of the left wall (y=0) and the ceiling (z=8). Let me consider the ceiling as extending from the left wall. If I unfold the left wall and ceiling into a plane, then the coordinates of the gecko on the left wall are (x, z), and on the ceiling, they would be (x, y) where y is the distance from the left wall. Wait, if the ceiling is unfolded adjacent to the left wall, then moving from the left wall to the ceiling would be moving in the y-direction. So, if the gecko walks from the left wall to the ceiling, then on the unfolded net, the gecko's path would be a straight line from its position on the left wall to some point on the ceiling, and then to the fly's position on the right wall.

But since the fly is on the right wall (y=10), which is connected to the ceiling, we need to figure out how to represent the fly's position on this unfolded net.

Alternatively, think of the entire path from the gecko's starting point to the fly's position as crossing the left wall, ceiling, and right wall. To represent this as a straight line, we need to have all three faces unfolded into a plane. However, due to the complexity, perhaps it's easier to use the method of reflecting the room.

Wait, here's a trick: if you imagine reflecting the room across the ceiling or floor, so that the gecko's path can be represented as a straight line across multiple reflections. For example, reflecting the fly's position across the ceiling to find a straight line path.

But since the gecko is allowed to walk on the ceiling and walls, the shortest path can be found by finding a straight line in the unfolded net where the gecko's start and end points lie on adjacent faces.

Alternatively, think about the following: If we unfold the left wall and the ceiling such that the right wall is adjacent to the ceiling, then the total distance can be calculated as follows:

The gecko starts at (11, 0, 7). If we unfold the left wall and the ceiling, the coordinates on the left wall are (x, z), and on the ceiling, they are (x, y). However, when we unfold the ceiling, so that it's adjacent to the left wall, we can imagine that the ceiling is now in the y-direction. So if we consider the ceiling unfolded, then the gecko can go from (x, z) to (x, y). But the fly is on the right wall (y=10). So perhaps if we unfold the ceiling and the right wall, then the fly is on the right wall, which is connected to the ceiling.

Alternatively, here's an idea: imagine that the gecko travels from the left wall to the ceiling to the right wall. If we unfold the left wall and the ceiling into a single plane, then the fly's position on the right wall can be represented as a point on the ceiling. Wait, how?

If we consider that the gecko travels from the left wall to the ceiling to the right wall, then on the unfolded net (left wall + ceiling), the fly's position is actually on the right wall, which is connected to the ceiling. So, if we imagine the right wall as being folded out from the ceiling, then on the unfolded net, the right wall would be adjacent to the ceiling, forming a larger plane.

Therefore, the coordinates on the unfolded net would be: for the left wall, (x, z), and on the ceiling and right wall, (x, y). Therefore, the total dimensions are: in the x-direction, it's 12 feet, and in the y-z combined direction? But how to map this.

Alternatively, think of it as follows: when unfolding the left wall, ceiling, and right wall into a single plane, the total width in one direction is the length of the room (12 feet), and the other direction is the height of the left/right wall (8 feet) plus the width of the ceiling (10 feet). Therefore, the total distance would be calculated as a straight line across this combined width.

But how exactly?

Wait, suppose we have the left wall (12x8) and the ceiling (12x10) and the right wall (12x8). If we unfold them so that they are all in a straight line along the y-direction, then the total length would be 8 (height of left wall) + 10 (width of ceiling) + 8 (height of right wall) = 26 feet? But that seems too long. Wait, no. If you lay them out side by side, the total length would be the sum of their widths in the respective directions. This is confusing.

Let me try to think of it in terms of coordinates again. If the gecko is on the left wall at (11, 0, 7), and the fly is on the right wall at (1, 10, 1). Suppose we unfold the left wall and the ceiling so that they form a plane. Then, the gecko's position on the left wall can be represented as (x, z), and the fly's position on the right wall is on the ceiling. Wait, no, the fly is on the right wall, not the ceiling.

Alternatively, if we go from the left wall to the ceiling, then to the right wall, then we can consider the ceiling as a bridge between them. So, on the ceiling, the fly is at (x=1, z=1), but since it's on the right wall, which is y=10, how does that translate?

Wait, if the gecko is on the left wall at (11, 0, 7), and it goes to the ceiling, which is at z=8. The ceiling is at z=8, and the fly is on the right wall at (1, 10, 1). So if we consider the gecko crossing from the left wall to the ceiling, then from the ceiling to the right wall, the coordinates on the ceiling would be (x, y). So, when on the ceiling, the gecko's path would be from (x1, y1) to (x2, y2). But since the gecko is moving from the left wall to the ceiling, which is adjacent, we can imagine that the starting point on the left wall corresponds to a point on the ceiling.

Alternatively, think of the left wall and ceiling as a single plane where the ceiling is the extension of the left wall's top. So, for example, if we have the left wall's top edge (z=8) connected to the ceiling's bottom edge (z=8). Then, if we imagine "flattening" the ceiling so that it's on the same plane as the left wall, then the coordinates on the ceiling would be (x, y) where y is measured from the left wall. Wait, that might work.

Therefore, on the left wall, the gecko is at (x=11, z=7). If we unfold the ceiling to be adjacent, we can represent the gecko's position as (x=11, z=7), and the fly is on the right wall (y=10). To reach the fly, the gecko would need to go from (11, 7) on the left wall to some point on the ceiling, then to the fly's position on the right wall. But since the fly is on the right wall, which is connected to the ceiling, we need to represent that on the unfolded net.

If we imagine that the ceiling is now extending in the y-direction from the left wall's top edge, then the fly's position on the right wall (which is at y=10) would be at a distance of y=10 from the left wall. However, since the ceiling is at y=0 to y=10, and connected to the left wall, the fly's position is at y=10 on the right wall. Wait, but the right wall is also at y=10. If we have the ceiling unfolded adjacent to the left wall, then the right wall is not directly connected. So perhaps we need to also unfold the right wall.

This is getting too complicated. Maybe I need to use the method of reflecting points across walls.

For example, to find the shortest path that goes from the left wall to the ceiling to the right wall, we can reflect the fly's position across the ceiling to the other side, and then find the straight line from the gecko's position to the reflection.

Wait, here's a method I remember: for problems involving shortest paths on the surface of a rectangular prism, you can reflect the room across the faces and find the straight-line distance between the original point and the reflected point.

For instance, if the gecko is moving from the left wall to the ceiling to the right wall, we can reflect the fly's position over the ceiling, but I need to think carefully.

Alternatively, consider that when moving from the left wall to the ceiling to the right wall, it's equivalent to going through two faces. So, imagine unfolding those two faces into a plane. For example, if we unfold the left wall and the ceiling, then the fly's position is on the right wall, which is a different face. Therefore, we might need to reflect the fly's position across the ceiling to a new position, and then calculate the distance from the gecko's original position to this reflection.

But I'm not sure. Let me try to find an example.

Suppose the gecko is on the left wall at (11, 0, 7), and the fly is on the right wall at (1, 10, 1). Let me consider unfolding the left wall and the ceiling. When we unfold them, the coordinates on the ceiling would be such that moving from the left wall to the ceiling is like moving along the y-axis. However, since the fly is on the right wall, which is at y=10, which is the end of the ceiling's width (y=10), maybe we can think of the fly's position on the ceiling as (1, 10, 1)? But no, because the ceiling is at z=8, so the fly is on the right wall, not the ceiling.

This is really challenging. Maybe I should look for another approach.

Let me think of the room as a rectangular prism with length=12, width=10, height=8. The gecko is at (11, 0, 7) on the left wall, and the fly is at (1, 10, 1) on the right wall. I need to find the shortest path along the surfaces.

One way to do this is to consider different possible unfoldings and calculate the distance. Let's try to think of possible unfoldings:

Unfolding 1: Left wall + ceiling + right wall. If we unfold the left wall, ceiling, and right wall so that they form a flat plane. This would create a large rectangle. Let's imagine that the left wall is on the left, the ceiling in the middle, and the right wall on the right. When unfolded, the total horizontal length is the same as the room's length (12 feet), and the vertical dimension is the height of the left wall (8 feet) plus the width of the ceiling (10 feet) plus the height of the right wall (8 feet)? No, that might not be correct.

Actually, if you unfold the left wall, ceiling, and right wall in sequence, the vertical dimension would be the height of the left wall plus the width of the ceiling? No, since the ceiling is horizontal. Wait, let's think about the dimensions:

- The left wall has dimensions 12 (length) x 8 (height).
- The ceiling has dimensions 12 (length) x 10 (width).
- The right wall has dimensions 12 (length) x 8 (height).

When we unfold the left wall, ceiling, and right wall in sequence, the total area would be a combination of these. However, to flatten them, the dimensions would be:

If you lay the left wall, then the ceiling, then the right wall next to each other, the total length in the direction of the length of the room (x-axis) is still 12 feet. The width would be the height of the left wall (8) + width of ceiling (10) + height of right wall (8)? No, that would be if they are stacked vertically, but since the ceiling is a horizontal surface, it's more like attaching it to the left wall's top, and then attaching the right wall to the end of the ceiling. So the combined dimensions would be:

- In the x-direction: 12 feet (same as room length).
- In the y-direction: 8 feet (height of left wall) + 10 feet (width of ceiling) + 8 feet (height of right wall)? No, that seems wrong.

Wait, actually, when you attach the ceiling to the left wall's top edge, the ceiling extends in the y-direction (width of the room), and the right wall is attached to the ceiling's end in the y-direction. So, the total dimension in the y-direction would be the height of the left wall (8) plus the width of the ceiling (10)? Because the left wall is vertical (height) and the ceiling is horizontal (width). If we unfold them such that the left wall is on the left, and the ceiling extends to the right from the top of the left wall, then the right wall would be attached to the end of the ceiling.

In this case, the total vertical dimension from the left wall's bottom to the right wall's top would be 8 (left wall height) + 10 (ceiling width) + 8 (right wall height)? No, that doesn't make sense. Let me think of it as two dimensions. If we have the left wall (12x8) unfolded with the ceiling (12x10), and then the right wall (12x8) added, the combined shape would have dimensions of 12 feet (length) by (8 + 10 + 8) feet? No, because the ceiling is attached to the top of the left wall, so it's not adding to the same dimension. This is not working.

Maybe I need to use coordinates. If I unfold the left wall and the ceiling such that the ceiling is attached to the top of the left wall, then the gecko's starting position is on the left wall at (x=11, z=7). The fly is on the right wall at (x=1, y=10, z=1). To represent this on the unfolded net, I need to find a coordinate system where both the gecko's position and the fly's position can be represented.

If I unfold the left wall and the ceiling, then the ceiling is in the y-direction from y=0 to y=10, and the left wall is in the z-direction from z=0 to z=8. But how do these combine?

Maybe if we consider the ceiling as extending in the y-direction from the left wall, then any point on the ceiling can be represented as (x, y), where y is from 0 to 10, and x is from 0 to 12. Meanwhile, the left wall is (x, z) where z is from 0 to 8.

To find the shortest path from (11, z=7) on the left wall to (1, y=10, z=1) on the right wall, we can imagine that the path goes from the left wall to the ceiling to the right wall. To model this on the unfolded net, we could reflect the fly's position across the ceiling.

Wait, if we reflect the fly's position across the ceiling, but I need to think.

Alternatively, think of the ceiling as a bridge. The gecko goes from left wall to ceiling to right wall. The straight-line distance on this unfolded net would be the hypotenuse of a right triangle, where one leg is the horizontal distance between the left wall and right wall, and the other is the vertical distance.

Wait, the horizontal distance between the left wall and the right wall is the length of the room, 12 feet, but since the gecko is one foot from the back wall (x=11) and the fly is one foot from the front wall (x=1), the distance along the x-axis is 11 - 1 = 10 feet? Or is it 12 - 1 - 1 = 10 feet?

Wait, the gecko is at x=11 on the left wall, and the fly is at x=1 on the right wall. So the distance along the x-axis between them is 11 - 1 = 10 feet? But since they are on opposite sides, it's actually 12 - 1 - 1 = 10 feet? Yes, because from x=1 (front) to x=11 (back) is 10 feet.

However, the path goes across the ceiling, so the vertical distance would involve the gecko's height and the fly's height.

Wait, the gecko is at z=7 (one foot from ceiling), and the fly is at z=1 (one foot from floor). If the gecko crosses the ceiling, it would go from z=7 to z=8 (ceiling) and then to z=1 on the right wall. But how does this translate in terms of distance?

Alternatively, if we consider the vertical distance as the difference in heights plus the height of the ceiling? No, this is confusing.

Let me try to think of it as a 2D problem by unfolding the left wall and ceiling into a plane. Let's say we have the left wall (x from 0 to 12, z from 0 to 8) and the ceiling (x from 0 to 12, y from 0 to 10). If we unfold the ceiling so that it is adjacent to the top edge of the left wall, then the combined surface is a rectangle with length 12 feet (x-axis) and height 8 + 10 = 18 feet (z + y). Wait, is that correct?

Yes! If you imagine unfolding the ceiling so that it is attached to the top of the left wall, then the total vertical dimension becomes the height of the left wall (8 feet) plus the width of the ceiling (10 feet). So in this unfolded net, the gecko's starting position is at (x=11, z=7) on the left wall. The fly is on the right wall, which is at y=10, but how does that translate?

Wait, if the ceiling is unfolded, then the right wall is connected to the end of the ceiling. So, if we imagine the unfolded net as a rectangle of 12 (x) by 18 (z+y), then the right wall is at y=10, which is 10 feet from the left wall (y=0). However, in this unfolded net, the coordinates would be such that moving from the left wall to the ceiling is moving in the y-direction. Therefore, the fly's position on the right wall is at (x=1, y=10, z=1). Wait, but on the unfolded net, the right wall is part of the ceiling?

No, the right wall is a separate face. If we have unfolded the left wall and the ceiling, the right wall is still separate. So to include the right wall, we need to unfold it as well.

Alternatively, if we include the right wall in the unfolding, then the total unfolded net would be the left wall, ceiling, and right wall, forming a larger rectangle. The left wall has width 12 (x) and height 8 (z). The ceiling has width 12 (x) and height 10 (y). The right wall has width 12 (x) and height 8 (z). If we unfold them such that they are connected along the x-axis, the total height would be 8 (left wall) + 10 (ceiling) + 8 (right wall)? No, because the right wall is at y=10, which is the same as the ceiling's width.

Actually, if you unfold the left wall, then the ceiling, then the right wall, all connected along the x-axis, the total height would be the height of the left wall (8) plus the width of the ceiling (10) plus the height of the right wall (8)? No, that would be if they are stacked vertically, but since the ceiling is horizontal, it's more like extending in the y-direction.

This is really too confusing without a diagram. Let me try to look for another approach.

Let me consider that the shortest path will cross two adjacent faces. For example, the left wall and the ceiling, or the left wall and the floor, etc. Let's calculate the distance for each case.

Case 1: Left wall to ceiling to right wall.

To calculate this, imagine that the gecko travels from its position on the left wall, across the ceiling, to the fly's position on the right wall. If we unfold the left wall and the ceiling into a single plane, the gecko's starting point is on the left wall, and the fly's position is on the right wall. However, since the right wall is connected to the ceiling, we need to adjust coordinates.

Let me define the unfolded net as follows: the left wall is a vertical rectangle, and the ceiling is a horizontal rectangle attached to the top of the left wall. In this net, the gecko's position is at (x=11, z=7), and the fly's position is on the right wall, which is at the end of the ceiling. But how to represent the fly's position?

The fly is on the right wall at (x=1, y=10, z=1). On the unfolded net, the right wall is not directly connected to the ceiling; it's connected to the ceiling's end in the y-direction. So if we consider the ceiling as extending in the y-direction, then the fly's position on the right wall is at y=10, x=1, z=1. Wait, but in the unfolded net, if we have the left wall and ceiling, the right wall is not included. So to include the right wall, we need to unfold it too.

Alternatively, imagine that the ceiling is connected to the right wall, so the total unfolded net is left wall + ceiling + right wall. In this case, the gecko's position is on the left wall at (11, 7), and the fly's position is on the right wall at (1, 10, 1). To find the straight-line distance on this unfolded net, we need to represent both positions in the same coordinate system.

In this unfolded net, the left wall is a rectangle of 12 x 8, the ceiling is 12 x 10, and the right wall is 12 x 8. If we lay them out such that the left wall is on the left, the ceiling is above it, and the right wall is on the right of the ceiling, then the coordinates would be:

- For the left wall: x from 0 to 12, z from 0 to 8.
- For the ceiling: x from 0 to 12, y from 0 to 10 (but attached to the top of the left wall).
- For the right wall: x from 0 to 12, z from 0 to 8 (attached to the end of the ceiling in the y-direction).

But this is still unclear. Let me try to assign coordinates in this unfolded net.

Let me imagine that in this net, we have:

- The left wall is the vertical part from (0, 0) to (12, 8) in a coordinate system where x is the length (0-12) and y is the height (0-8).

- The ceiling is attached to the top of the left wall, extending in the y-direction from (0, 8) to (12, 8 + 10 = 18). So the ceiling is a horizontal strip from (0, 8) to (12, 18).

- The right wall is attached to the end of the ceiling, extending in the z-direction from (12, 8) to (12 + 8 = 20, 18)? No, this is getting too convoluted.

Alternatively, let me think of the combined unfolded net as having the left wall, ceiling, and right wall forming a rectangle with length 12 (x-axis) and height 8 + 10 = 18 (z + y). Wait, if I consider the left wall (z=0 to 8) and the ceiling (y=0 to 10), then if I unfold them, the total height would be 8 + 10 = 18. But the right wall is at y=10, so in this unfolded net, the fly's position is at (x=1, y=10, z=1)? No, I need to relate it.

Wait, if the gecko is on the left wall at (x=11, z=7), and in the unfolded net, the left wall is from x=0 to 12, z=0 to 8, then the gecko's position is at (11, 7). The fly is on the right wall at (x=1, y=10, z=1). But in the unfolded net, the right wall is at y=10, so if we consider the unfolded net as z from 0 to 8 (left wall) and y from 0 to 10 (ceiling), then the fly's position would be at (x=1, y=10, z=1)? But how does z=1 relate to this?

This is really too confusing. Maybe I need to use a different method.

Let me try using the method of unfolding the room such that the path crosses two adjacent faces. For example, if the gecko goes from the left wall to the ceiling to the right wall, then the total horizontal distance would be the distance between the gecko and the fly along the x-axis, and the total vertical distance would be the distance from the gecko's height to the fly's height plus the height of the ceiling?

Wait, the gecko is one foot from the ceiling (z=7), and the fly is one foot from the floor (z=1). If the gecko crosses the ceiling, it would go from z=7 to z=8 (ceiling) and then to z=1 on the right wall. So the vertical distance would be (8 - 7) + (8 - 1) = 1 + 7 = 8? Wait, no, that's if it goes up and down. But in terms of the straight line on the unfolded net, it would be the distance between the gecko's position and the fly's position, accounting for the unfolded dimensions.

Let me imagine that by unfolding the left wall and the ceiling, the gecko's position is at (11, 7) in a coordinate system where x is length and z is height, and the fly's position is somewhere on the ceiling. But the fly is on the right wall, not the ceiling, so perhaps I need to also include the right wall in the unfolding.

Alternatively, think of the gecko's path as going from (11, 0, 7) to (1, 10, 1). If we unfold the left wall and the ceiling, then the ceiling is 12 x 10, and the fly is on the right wall. If we also unfold the right wall, then the total dimensions would be such that the gecko's path is a straight line across the left wall, ceiling, and right wall.

But without knowing the exact unfolding, it's hard to proceed. Let me try to think of it as a 3D problem and project it onto 2D.

Let me consider the gecko's path as going from the left wall to the ceiling to the right wall. If I imagine the ceiling as a flat surface, the gecko's path would be a straight line from its position on the left wall to the fly's position on the right wall, passing through the ceiling. To calculate this, we can think of the ceiling as a rectangle and calculate the distance as follows:

The gecko is 11 feet from the front (x=11), 7 feet from the floor (z=7). The fly is 1 foot from the front (x=1), 1 foot from the floor (z=1). If the gecko crosses the ceiling, which is at z=8, then the vertical distance from the gecko to the ceiling is 8 - 7 = 1 foot, and from the ceiling to the fly is 8 - 1 = 7 feet. But this is in 3D, and we need to convert it to 2D.

Wait, if we unfold the left wall and the ceiling into a plane, then the gecko's position is at (x=11, z=7) and the fly's position, when projected onto the ceiling, would be at (x=1, z=8) because it's on the ceiling? No, the fly is on the right wall, not the ceiling. Wait, if we go from the left wall to the ceiling to the right wall, then on the ceiling, the gecko would reach a point that is on the way to the right wall. Let me think of it as two separate steps: from left wall to ceiling, and then from ceiling to right wall.

Let me calculate the distance from the gecko to a point on the ceiling, and from there to the fly. Let's call the point where the path meets the ceiling as (x, 8) on the ceiling. Then the total distance is the distance from (11, 0, 7) to (x, 8) on the ceiling, plus the distance from (x, 8) to (1, 10, 1) on the right wall.

But this is a minimization problem, where we need to find x that minimizes the total distance. However, this might be complex. Alternatively, since the shortest path should be a straight line in the unfolded net, we can find x such that the path is straight.

If we unfold the left wall and the ceiling into a single plane, then the gecko's position is (11, 7) on the left wall, and the fly's position on the right wall would be at (1, 10, 1). But how to represent this in the unfolded net?

Wait, if we unfold the left wall and the ceiling, the ceiling is now a horizontal surface. The right wall is at y=10, so if we imagine the ceiling extending in the y-direction, then the fly's position on the right wall is at (x=1, y=10, z=1). But on the unfolded net, this would be at (x=1, y=10) on the ceiling? Wait, no, because the fly is on the right wall, not the ceiling.

This is getting too tangled. Maybe I should look for an example of similar problems.

In many shortest path problems on a rectangular prism, the shortest path can be found by unfolding the prism into a flat net and measuring the straight line distance. For example, if the gecko needs to go from one point to another across two adjacent faces, the distance is the hypotenuse of a right triangle with sides equal to the sum of certain dimensions and the difference of others.

Let me try to apply this. Suppose the gecko goes from the left wall to the ceiling to the right wall. The total distance would be the hypotenuse of a right triangle where one leg is the distance along the length of the room (from x=11 to x=1, which is 10 feet) and the other leg is the distance from the gecko's height to the fly's height plus the height of the ceiling? Or something like that.

Wait, if we consider the gecko going from (11, 0, 7) to the ceiling, then to (1, 10, 1), the total horizontal distance would be from x=11 to x=1, which is 10 feet. The vertical distance would be from z=7 to z=8 (1 foot up) and then from z=8 to z=1 (7 feet down), totaling 8 feet. So the total distance would be sqrt(10² + 8²) = sqrt(164) ≈ 12.8 feet. But this is just a guess.

Alternatively, if the unwrapped net is such that the total horizontal distance is 10 feet and the total vertical distance is 7 (from z=7 to z=1) plus 1 (from z=7 to ceiling)? No, I'm not sure.

Wait, let's think of it this way: if the gecko crosses the ceiling, the path would form a rectangle when unfolded. The left wall and the ceiling form a rectangle of length 12 (x) and height 8 (z) + 10 (y)? No, it's more like the left wall (12 x 8) and the ceiling (12 x 10) combined. If we unfold them, the total height would be 8 + 10 = 18, and the length is 12. But the gecko's position is at (11, 7) on the left wall, and the fly is at (1, 10, 1) on the right wall. Wait, but where is the fly in this unfolded net?

If we imagine the left wall and the ceiling as part of the net, the right wall is not included. So perhaps we need to include it as well. If we unfold the left wall, ceiling, and right wall, the total height is 8 (left wall) + 10 (ceiling) + 8 (right wall) = 26? No, this is incorrect.

Alternatively, if the gecko goes from the left wall to the ceiling to the right wall, the total distance would be the hypotenuse of a right triangle with one leg being the length from x=11 to x=1 (10 feet) and the other leg being the height difference from z=7 to z=1 (6 feet) plus the height of the ceiling (8 feet)? No, this is not coherent.

Wait, perhaps the correct way is to think of the path as going from (11, 0, 7) to (1, 10, 1) through the ceiling. If we unfold the left wall and the ceiling into a plane, then the coordinates on this plane would be such that the ceiling extends in one direction. Let me assign coordinates: let's say the left wall is in the x-z plane, and the ceiling is in the x-y plane. If we unfold them such that the x-axis is common, then the total distance would be calculated by combining the z and y dimensions.

So, the gecko's position is at (11, 7) in x-z, and the fly's position is at (1, 10) in x-y, but they are on different walls. Wait, no, the fly is on the right wall, which is at y=10. If the ceiling is unfolded, then the right wall is at y=10, z=8? Not sure.

Alternatively, think of the total distance as a rectangle where the gecko travels from (11, 7) to (1, 10) in a combined x-y and x-z coordinate system. This is too vague.

At this point, I think I need to refer to the standard method for solving such problems, which is to consider the room as a box and find the shortest path by unfolding it into a plane.

For example, the shortest path from one point to another on a box can be found by considering different possible nets and calculating the distance for each.

Let me try to consider the path going across the left wall and the ceiling. To do this, the gecko would travel from the left wall to the ceiling, and then to the right wall. When you unfold the left wall and the ceiling into a single plane, the fly's position on the right wall is now on the ceiling. Wait, how?

If we unfold the left wall and the ceiling, then the right wall is adjacent to the ceiling. So, the fly's position on the right wall is at (1, 10, 1), but in this unfolded net, it would be at (1, 10) on the right wall, which is connected to the ceiling. If we represent this in the unfolded net, the total distance would be the straight line from (11, 7) on the left wall to (1, 10) on the right wall, with the ceiling in between.

But how to calculate this? If we imagine the left wall and ceiling as a single plane, the total height would be the height of the left wall (8) plus the width of the ceiling (10)? No, if the ceiling is attached to the top of the left wall, then the total height is 8 + 10 = 18, but the x-axis is still 12.

Wait, let me define the unfolded net as having the left wall (12x8) and the ceiling (12x10) attached to it. So the combined surface is a rectangle of 12x18. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10 + 8 = 18)? Wait, no.

Actually, if the ceiling is attached to the top of the left wall, then the ceiling's y-coordinate (which is 10) would be added to the left wall's z-coordinate (8). So, the fly's position on the right wall is at x=1, y=10, z=1. But in the unfolded net, the right wall is at y=10, so if we consider the ceiling to be in the y-direction, then the fly's position is at y=10, x=1, and z=1? This is confusing.

Maybe it's better to use the following method: when you unfold the left wall and the ceiling, the total vertical dimension is the height of the left wall (8) plus the width of the ceiling (10). The gecko's position is at (x=11, z=7) on the left wall, and the fly's position is at (x=1, y=10, z=1) on the right wall. To represent the fly's position in this unfolded net, we need to map it to the combined dimensions.

If we consider the left wall's z-axis and the ceiling's y-axis as a single dimension, then the fly's position would be at a distance of 8 (height of left wall) + 10 (width of ceiling) - 1 (z=1)? No, this is not linear.

Let me try to think of it as a 2D coordinate system where one axis is the length of the room (x from 0 to 12) and the other axis is a combined dimension of height (z) and width (y). If the gecko goes from the left wall to the ceiling, then the combined dimension would be z (from 0 to 8) + y (from 0 to 10). So, the gecko's position is at (x=11, z=7), and the fly's position is at (x=1, y=10, z=1). In this combined system, the z and y are combined into a single dimension. But how?

If we represent the combined dimension as 8 + 10 = 18, then the gecko's position is at (11, 7) and the fly's position is at (1, 10 + 1 = 11)? Wait, why 10 + 1? If the fly is on the right wall at z=1, then in the combined dimension, it would be 8 (height of left wall) + 1 (z=1 from the floor)? No, this is not correct.

I think I'm stuck here. Let me try to look for another approach.

Let me recall that in similar problems, the shortest path can be found by considering the room as a box and "flattening" it into a net. For example, if the gecko goes from the left wall to the ceiling to the right wall, the distance can be calculated as follows:

The gecko travels across the left wall and the ceiling. The total distance would be the hypotenuse of a right triangle where one leg is the length of the room minus the distances from the start and end points, and the other leg is the sum of the heights.

Wait, let me try this. The gecko is 1 foot from the back wall (x=11), and the fly is 1 foot from the front wall (x=1). The length between them is 11 - 1 = 10 feet. The gecko is 1 foot from the ceiling (z=7), and the fly is 1 foot from the floor (z=1). If the path goes across the ceiling, the vertical distance would be from z=7 to z=8 (1 foot) and from z=8 to z=1 (7 feet), totaling 8 feet. So the distance would be sqrt(10² + 8²) = sqrt(164) = 2√41 ≈ 12.806 feet.

But is this correct? Let me check.

If the gecko goes from (11, 0, 7) to the ceiling at (11, 0, 8), which is 1 foot, and then from there to (1, 10, 1), which would be a distance of sqrt((11-1)^2 + (0-10)^2 + (8-1)^2) = sqrt(100 + 100 + 49) = sqrt(249) ≈ 15.78 feet, which is longer. So this approach is incorrect.

Wait, but if we unfold the left wall and the ceiling, then the path is a straight line. So let me think of it as a 2D problem where the left wall and ceiling are flat. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10) on the right wall. But how?

If we imagine the left wall and ceiling as a single plane, then the right wall is at a distance of 10 feet (width of the room) from the left wall. Wait, no. The width of the room is 10 feet, but the gecko is on the left wall (y=0), and the fly is on the right wall (y=10). So the distance between the left and right walls is 10 feet.

If we unfold the left wall and the ceiling, then the gecko's position is at (x=11, z=7), and the fly's position is at (x=1, y=10, z=1). But if the ceiling is unfolded, the y-coordinate is now part of the vertical dimension. So the total vertical distance is the z-coordinate of the gecko (7) plus the y-coordinate of the fly (10) minus the z-coordinate of the fly (1)? No, this is not clear.

At this point, I think I need to accept that I'm struggling with visualizing the net and look for a different strategy.

Let me recall that when dealing with shortest paths on a cube, the shortest path can often be found by considering different possible unfoldings and calculating the Euclidean distance.

For this problem, the room is a rectangular prism, not a cube, but similar principles apply.

Let me consider the following: the gecko is on one side wall, and the fly is on the opposite side wall. To get from one to the other, the gecko can go across the ceiling, floor, or a combination of other walls.

Let me calculate the distance for the path that goes from the left wall across the ceiling to the right wall, and similarly for the floor.

For the ceiling path: If we imagine unfolding the left wall and the ceiling into a single plane, the gecko's position is at (11, 7) and the fly's position is at (1, 10). Wait, how?

Let me think of the combined height as the height of the left wall (8) + width of the ceiling (10) = 18. But the gecko is at z=7 (7 feet from the floor) on the left wall, and the fly is at 1 foot from the floor on the right wall, which is 1 foot from the floor. If we imagine the ceiling as extending in the y-direction, then the fly's position is at y=10 (width of the room) and z=1 (height from floor). But in the unfolded net, this would be at a distance of 8 (height of left wall) + 1 (height from floor on the right wall) = 9? No, this is not correct.

Wait, if we unfold the left wall and the ceiling, the height from the floor on the left wall is z=7, and on the ceiling, it would be y=10 (width) and z=8 (height). But the fly is on the right wall at z=1, which is 1 foot from the floor. If we imagine the right wall as being attached to the ceiling, then the fly's position would be at y=10 (width) and z=1 (height from floor), which in the unfolded net would be at y=10 and z=1, but how does that relate to the left wall's z=7?

Maybe I should think of it as follows: when you unfold the left wall and ceiling, the total vertical dimension is 8 (left wall height) + 10 (ceiling width) = 18. The gecko is 7 feet up on the left wall, and the fly is 1 foot up on the right wall. If the right wall is at the end of the ceiling, then the fly's position is at 10 feet in the width direction (y=10) and 1 foot in the height (z=1). But how do these combine?

If we consider the combined vertical dimension as z (from 0 to 8) and y (from 0 to 10), then the gecko's position is at (11, 7) and the fly's position is at (1, 10, 1). Wait, this is 3D, but if we flatten it into 2D, it would be a rectangle of (x, z + y)? No, this is too abstract.

Let me try to use the formula for the shortest path on a rectangular prism. The shortest path can be calculated by finding the minimal distance across different nets.

For example, if the gecko travels across the left wall and the ceiling, the distance can be calculated as follows:

The path would go from (11, 0, 7) to (1, 10, 1) via the ceiling. If we unfold the left wall and the ceiling, the total distance is the hypotenuse of a right triangle with one leg being the length between x=11 and x=1 (10 feet), and the other leg being the vertical distance from z=7 to z=8 (1 foot) plus the vertical distance from z=8 to z=1 (7 feet), totaling 8 feet. So the distance is sqrt(10² + 8²) = sqrt(164) = 2√41 ≈ 12.806.

Alternatively, if the gecko goes across the left wall and the floor, the vertical distance would be from z=7 to z=0 (7 feet) and from z=0 to z=1 (1 foot), totaling 8 feet. So the distance is also sqrt(10² + 8²) = sqrt(164). So both paths give the same distance.

But is there a shorter path through other faces?

Let me consider going from the left wall to the front wall to the right wall.

The front wall is at x=0. The gecko is at x=11 on the left wall, so the distance along the x-axis from the gecko to the front wall is 11 feet. The front wall is 10 feet wide (y) and 8 feet tall (z). The fly is at x=1, y=10, z=1. If the gecko goes from the left wall to the front wall to the right wall, the total distance would be the hypotenuse of a right triangle with one leg being the distance along the x-axis from 11 to 0 (11 feet) and the other leg being the distance from the gecko's position to the fly's position in y and z. But this is getting complex.

Let me try to calculate it. If the gecko goes from (11, 0, 7) to the front wall, then to (1, 10, 1). If we unfold the left wall and the front wall, the total distance would be the hypotenuse of a right triangle with one leg being the distance from x=11 to x=1 (10 feet) and the other leg being the distance from z=7 to z=1 (6 feet) plus the width of the front wall? No, this is not clear.

Alternatively, if the gecko goes from the left wall to the front wall (11 feet in x-direction), and then from the front wall to the right wall (1 foot in x-direction), but this is not helpful.

Alternatively, using the same method as before, if the gecko goes from left wall to front wall to right wall, the total distance would be the hypotenuse of a right triangle with one leg being the length from x=11 to x=1 (10 feet) and the other leg being the sum of the height from z=7 to z=1 (6 feet) and the width from y=0 to y=10 (10 feet)? No, this is not correct.

At this point, I think the shortest path is likely the one that goes across the ceiling or floor, giving a distance of sqrt(10² + 8²) = sqrt(164) = 2√41. But I need to confirm.

Wait, let's calculate the distance if the gecko goes from the left wall to the ceiling to the right wall. If we consider the left wall and the ceiling as one plane, and the right wall as another, the distance would be the straight line from (11, 7) to (1, 10 + 8 - 1)? No, this is not coherent.

Wait, let me try to use the method of reflecting the fly's position across the ceiling. If the gecko goes from the left wall to the ceiling to the right wall, we can reflect the fly's position across the ceiling to find the straight line.

If the fly is at (1, 10, 1), its reflection across the ceiling (z=8) would be at (1, 10, 15). The distance from the gecko's position (11, 0, 7) to the reflected point (1, 10, 15) would be sqrt((11 - 1)^2 + (0 - 10)^2 + (7 - 15)^2) = sqrt(100 + 100 + 64) = sqrt(264) = 2√66 ≈ 16.248, which is longer than the previous calculation, so this is not helpful.

Alternatively, if we reflect the gecko's position across the ceiling, but this is getting too complex.

Wait, let me think of the problem as a 2D problem by considering the room's dimensions.

If the gecko goes from the left wall to the ceiling to the right wall, we can imagine flattening the left wall and ceiling into a plane. The total distance would be the hypotenuse of a right triangle with one leg being the distance between the x-coordinates (11 to 1 = 10 feet) and the other leg being the distance in the y-z plane. The y-z plane here would be the height of the gecko (7 feet) plus the height from the ceiling to the fly (1 foot), but this is not correct.

Wait, if the gecko is at (11, 0, 7) and the fly is at (1, 10, 1), and the path goes across the ceiling, then on the unfolded net, the distance would be from (11, 7) to (1, 10 + 8 - 1)? No, I'm not sure.

Let me try to think of the unfolded net as having the left wall and the ceiling combined. The left wall has dimensions 12 x 8, and the ceiling has 12 x 10. If we lay them flat, the total height would be 8 + 10 = 18, and the length is 12. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10) on the right wall. Wait, if the right wall is at the end of the ceiling, then the fly's position is at (1, 10) on the right wall, which is 10 feet in the y-direction (width) and 1 foot in the z-direction (height). But in the unfolded net, the total height is 18, so the fly's position would be at (1, 10 + 8 - 1) = (1, 17)? No, this is not logical.

I think I need to accept that I'm not able to visualize the correct unfolding and look for another way.

Let me consider that the shortest path will involve going across two adjacent walls. Let's calculate the distance for each possible pair of walls and choose the smallest.

Possible pairs:

1. Left wall and ceiling: distance = sqrt( (11 - 1)^2 + (7 + 1)^2 ) = sqrt(100 + 64) = sqrt(164)

Wait, how did I get (7 + 1)? If the gecko goes from z=7 on the left wall to z=8 on the ceiling (1 foot), and then from z=8 to z=1 on the right wall (7 feet), but in the unfolded net, it's a straight line. If the total vertical distance is 7 + 1 = 8, then the distance is sqrt(10² + 8²) = sqrt(164).

2. Left wall and floor: similar calculation, sqrt(10² + (7 + 1)^2) = sqrt(164).

3. Left wall and front wall: if the gecko goes from left wall to front wall to right wall. The front wall is at x=0, so the distance from x=11 to x=0 is 11 feet. The vertical distance would be from z=7 to z=1, which is 6 feet, and the width from y=0 to y=10 is 10 feet. But this would form a right triangle with legs 11 and 10 + 6 = 16? No, this is not correct.

Wait, if the gecko goes from left wall to front wall to right wall, it would traverse the left wall to front wall (which is 11 feet in x), and then front wall to right wall (1 foot in x). But the vertical distance would be from z=7 to z=1, which is 6 feet. So the total distance would be sqrt(11² + 6²) + sqrt(1² + 10²)? No, this is not correct.

Alternatively, if we unfold the left wall and front wall, the distance would be sqrt((11 + 1)^2 + (7 + 1)^2) = sqrt(144 + 64) = sqrt(208) = 4√13 ≈ 14.422, which is longer than sqrt(164) ≈ 12.806.

4. Left wall and back wall: the gecko is already one foot from the back wall (x=11), so going to the back wall would be 0 feet, but the fly is on the opposite wall, so this might not help.

5. Left wall and right wall: but they are opposite, so the gecko cannot go directly.

So, the shortest distance seems to be sqrt(164), which is 2√41. Let me check if there are other possible paths.

Wait, what if the gecko goes across the ceiling and the front wall? Let me try that.

If the gecko goes from left wall to ceiling to front wall to right wall, but this would be a longer path. Let me calculate it.

But this is getting too complex. Based on the previous calculations, the shortest path seems to be sqrt(164). Let me check if this is correct.

Wait, let me verify with another approach. Let's consider the room as a box with dimensions 12x10x8. The gecko is at (11, 0, 7) and the fly is at (1, 10, 1). The shortest path on the surfaces can be found by considering the net where the path crosses two adjacent faces.

If we consider the left wall (12x8) and the ceiling (12x10) as a single net, the distance would be the hypotenuse of a right triangle with one leg being the distance between x=11 and x=1 (10 feet) and the other leg being the distance between z=7 and z=1 (6 feet) plus the width of the ceiling (10 feet)? No, this is not correct.

Wait, another way: if we unfold the left wall and the ceiling such that the gecko's path is a straight line, the total vertical distance would be the distance from z=7 on the left wall to z=1 on the right wall, which is 7 - 1 = 6 feet, but since it goes through the ceiling, it would be 8 - 7 + 1 = 2 feet? No, this is confused.

At this point, I think I need to proceed with the sqrt(164) answer, which is 2√41, and see if it matches with the correct answer.

But let me check another possibility: if the gecko goes across the ceiling and the right wall, but I think it's the same as going across the ceiling.

Alternatively, let me calculate the distance if the gecko goes across the left wall and the ceiling, and the right wall is part of the ceiling. Wait, the right wall is at y=10, so if we imagine unfolding the ceiling and the right wall, the total width would be 10 (ceiling) + 8 (right wall) = 18. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 1) on the right wall. Wait, no.

If we unfold the ceiling and the right wall, the gecko's position on the left wall is now at (11, 7) on the left wall, and the fly is at (1, 1) on the right wall. The distance would be sqrt((11 - 1)^2 + (7 + 10 + 1)^2)? No, this is not correct.

I think I need to accept that the shortest path is 2√41, which is sqrt(164), and check if this is the correct answer.

But let me think again: if the gecko goes from (11, 0, 7) to (1, 10, 1) via the ceiling, the path would be a straight line in the unfolded net. If we consider the left wall and ceiling as a single plane, then the total vertical dimension is 8 (height of left wall) + 10 (width of ceiling) = 18. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10) on the right wall. Wait, if the right wall is at the end of the ceiling, then the fly's position is at x=1, y=10, which in the unfolded net would be at x=1, y=10 + 8 = 18? No.

Alternatively, if the fly is on the right wall at (1, 10, 1), and we're in the unfolded net of left wall and ceiling, then the fly's position is at (1, 10) on the right wall, which is at a distance of 10 feet in y and 1 foot in z. But how does this translate to the unfolded net?

I think I'm going in circles. Given that I've spent a lot of time and the most plausible answer seems to be sqrt(164) or 2√41, I'll go with that, but I need to check if there's a shorter path.

Wait, let me try another unfolding. Suppose the gecko goes from the left wall to the front wall to the right wall. If we unfold the left wall and the front wall, the total distance would be the hypotenuse of a right triangle with one leg being the distance from x=11 to x=1 (10 feet) and the other leg being the distance from z=7 to z=1 (6 feet) plus the width of the front wall (10 feet)? No, this is not correct.

Wait, if we unfold the left wall and the front wall into a single plane, the total dimensions would be 12 (length) by 8 (height) + 10 (width) = 18. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 10) on the right wall. The distance would be sqrt((11 - 1)^2 + (7 + 10 - 1)^2) = sqrt(100 + 16²) = sqrt(100 + 256) = sqrt(356) ≈ 18.87, which is longer.

Therefore, the shortest path is indeed sqrt(164) = 2√41.

But let me verify with another possible unfolding. What if the gecko goes across the ceiling and the right wall? Let me imagine unfolding the ceiling and the right wall. The total dimensions would be 12 (length) by 8 (height of right wall) + 10 (width of ceiling) = 18. The gecko's position is at (11, 7) on the left wall, which is now at (11, 8 + 7 - 8 = 7)? No, this is not clear.

Alternatively, if the gecko goes from the left wall to the ceiling to the right wall, the unfolded net is left wall (12x8) + ceiling (12x10) + right wall (12x8). If we lay them out in a straight line, the total height is 8 + 10 + 8 = 26. The gecko's position is at (11, 7) on the left wall, and the fly's position is at (1, 1) on the right wall. The distance would be sqrt((11 - 1)^2 + (7 + 10 + 1)^2) = sqrt(100 + 18²) = sqrt(100 + 324) = sqrt(424) ≈ 20.59, which is longer.

Thus, the shortest path is indeed sqrt(164), which simplifies to 2√41.

Wait, but let me check if there's a different unfolding where the distance is smaller. For example, what if the gecko goes across the ceiling and the back wall? Or other combinations?

Let me try another possibility: the gecko goes from the left wall to the back wall to the right wall. The back wall is at x=12. The gecko is at x=11, so it's 1 foot from the back wall. If it goes to the back wall, then to the right wall. The distance would be sqrt(1² + (7 + 1)^2) + sqrt(12 - 1)^2 + ... No, this is not helpful.

Alternatively, if we unfold the back wall and the right wall, but this is getting too complex.

Given all the above, I think the shortest path is 2√41, which is sqrt(164). Let me check if this is correct.

Wait, but let me calculate 10² + 8² = 100 + 64 = 164. Yes, sqrt(164) is approximately 12.806.

But I recall that in similar problems, sometimes the shortest path involves going across three faces, but in this case, it seems that going across two faces is sufficient.

Wait, but let me think: if the gecko goes from the left wall to the ceiling to the right wall, the actual distance is sqrt((11-1)^2 + (7 + 10 - 1)^2)? No, why 7 + 10 - 1?

Wait, if the gecko is at (11, 0, 7) on the left wall, and the fly is at (1, 10, 1) on the right wall, and the path goes across the ceiling. If we unfold the left wall and the ceiling, then the total vertical distance is from z=7 on the left wall to z=1 on the right wall, passing through the ceiling. If we consider the ceiling as extending in the y-direction, the vertical distance would be 7 (from z=7 to z=8) + 1 (from z=8 to z=1) = 8. So the distance is sqrt(10² + 8²) = sqrt(164). This seems to be the correct calculation.

Yes, this makes sense. So the shortest path is 2√41, which is the simplified radical form.

**Final Answer**
The length of the shortest path is \boxed{2\sqrt{41}}.

To find the shortest path a gecko can take from one point to another on the **surface** of a rectangular room, it's helpful to **unfold** the room into a 2D net. This allows us to calculate the straight-line distance (which represents the path along the surfaces of the room) using the **Pythagorean Theorem**.

---

### **1. Understanding the Room and Coordinates**

We are given:
- **Room dimensions**: 12 feet (length × x-axis), 10 feet (width × y-axis), and 8 feet (height × z-axis).
- **Gecko's starting position**: On the **left wall** (y = 0), one foot from the ceiling (z = 7) and one foot from the **back wall** (x = 11). So,
  - Start: (11, 0, 7)
- **Fly's position**: On the **opposite right wall** (y = 10), one foot from the **floor** (z = 1) and one foot from the **front wall** (x = 1). So,
  - End: (1, 10, 1)

---

### **2. Choosing the Shortest Path**

Since the gecko is on the **left wall** and the fly is on the **opposite right wall**, the shortest path likely involves **moving across two adjacent faces**—for example, the **left wall** and the **ceiling**. This way, the gecko travels across the **least amount of surface area** and avoids unnecessary detours.

If the gecko crosses the ceiling, the path becomes a straight line in a 2D unfolded net, where:
- The **horizontal distance** between the gecko and the fly is from x = 11 to x = 1:
  $$
  \text{Horizontal distance} = 11 - 1 = 10 \text{ feet}
  $$
- The **vertical distance** is from z = 7 (on the left wall) to z = 1 (on the right wall), passing through the ceiling. The total vertical distance is:
  $$
  \text{Vertical distance} = (8 - 7) + (8 - 1) = 1 + 7 = 8 \text{ feet}
  $$

Now, treat this as a right triangle with:
- Base: 10 feet
- Height: 8 feet

---

### **3. Computing the Shortest Path**

Using the **Pythagorean Theorem**:
$$
\text{Shortest path} = \sqrt{10^2 + 8^2} = \sqrt{100 + 64} = \sqrt{164}
$$

Simplify:
$$
\sqrt{164} = \sqrt{4 \times 41} = 2\sqrt{41}
$$

---

### **Final Result**

$$
\boxed{2\sqrt{41}}
$$

This is the **shortest possible path** the gecko can take on the surface of the room to reach the fly, assuming it cannot jump and is allowed to walk across the floor, ceiling, and walls.

Solution:
[asy]
import three;
currentprojection=orthographic(1/2,-1,1/2);
triple A,B,C,D,E,F,G,H,g,f;
A = (0,0,0);
B = (12,0,0);
C = (12,10,0);
D = (0,10,0);
E = (0,10,8);
F = (0,0,8);
G = (12,0,8);
H = (12,10,8);
draw(A--B--C--D--cycle);
draw(E--F--G--H--cycle);
draw(A--F); draw(B--G); draw(C--H); draw(D--E);
g = (12,9,7); f = (0,1,1);
dot(g, green); dot(f, purple);
label("12", A--B); label("10", B--C); label("8", C--H);
[/asy] In the above diagram, the green dot is the gecko and the purple dot is the fly. We can ``unfold'' the walls that the gecko traveled along, as below, to represent the gecko's path in two dimensions. This unfolding does not change the length of the gecko's path, so in order for the gecko's path to be minimal before unfolding, it must be minimal after unfolding. In other words, it must be a straight line after unfolding. Now, besides the side walls, the gecko can travel along the front, back, and ceiling. Suppose that among these, it only travels along the front wall. The walls the gecko walked along unfold as such: [asy]
draw( (0,0)--(10,0)--(10,8)--(0,8)--cycle ); draw( (10,0)--(22,0) ); draw( (10,8)--(22,8) );
draw( (22,0)--(32,0)--(32,8)--(22,8)--cycle );
pair g = (31,7); pair f = (9,1);
dot(g, green); dot(f, purple);
draw(g--f, red);
draw(f--(31,1), red+dashed); draw(g--(31,1), red+dashed);
label( "10", (0,0)--(10,0) ); label( "12", (10,0)--(22,0) ); label( "10", (22,0)--(32,0) ); label( "8", (32,0)--(32,8) );
[/asy] The gecko's path is the hypotenuse of a right triangle with legs 6 and 22, so its length is $\sqrt{6^2 + 22^2} = 2\sqrt{3^2 + 11^2} = 2\sqrt{130}$. By symmetry (the gecko and the fly are exactly opposite each other in the room), the path length is the same if the gecko only travels along the back wall and side walls.

Now suppose the gecko only travels along the ceiling and side walls. These walls unfolded become: [asy]
draw( (0,0)--(8,0)--(8,10)--(0,10)--cycle ); draw( (8,0)--(20,0) ); draw( (8,10)--(20,10) );
draw( (20,0)--(28,0)--(28,10)--(20,10)--cycle );
pair g = (21,9); pair f = (1,1);
dot(g, green); dot(f, purple);
draw(g--f, red);
draw(f--(21,1), red+dashed); draw(g--(21,1), red+dashed);
label( "8", (0,0)--(8,0) ); label( "12", (8,0)--(20,0) ); label( "8", (20,0)--(28,0) ); label( "10", (28,0)--(28,10) );
[/asy] The path is the hypotenuse of a right triangle with legs 8 and 20, so its length is $\sqrt{8^2 + 20^2} = 2\sqrt{4^2+10^2} = 2\sqrt{116}$. (We'll keep it in this form because it makes it easier to compare with the other cases.)

Finally, the gecko may cross both the ceiling and front wall (or back wall; the cases give the same results by symmetry). The unfolded walls then look like this: [asy]
draw( (0,0)--(10,0)--(10,8)--(0,8)--cycle );
draw( (10,0)--(22,0)--(22,8)--(10,8)--(10,18)--(22,18) );
draw( (22,8)--(30,8)--(30,18)--(22,18)--cycle );
pair g = (23,17); pair f = (9,1);
dot(g, green); dot(f, purple);
draw(g--f, red);
draw(f--(23,1), red+dashed); draw(g--(23,1), red+dashed);
label("10", (0,0)--(10,0)); label("12", (10,0)--(22,0)); label("8", (0,0)--(0,8), W);

label("8", (22,18)--(30,18), N); label("10", (30,18)--(30,8), E);
[/asy] The path is the hypotenuse of a right triangle with legs 16 and 14, so its length is $\sqrt{16^2+14^2} = 2\sqrt{8^2+7^2} = 2\sqrt{113}$. Of the three cases, this is the smallest, so the answer is $\boxed{2\sqrt{113}}$.