Explanation:
Okay, so I need to find the square of the determinant of a 3x3 matrix with columns (x1, y1, 1), (x2, y2, 1), and (x3, y3, 1). The determinant squared is what's being asked. The problem gives me three equations involving the distances between three points (x1,y1), (x2,y2), and (x3,y3). Specifically, the distances squared between each pair are 9, 16, and 25.

First, let me recall that the determinant of a 3x3 matrix with columns as points (x1,y1,1), etc., relates to the area of the triangle formed by those three points. Wait, actually, the determinant itself is twice the area of the triangle, right? Because the area can be calculated using the shoelace formula, and the determinant is related to that. But since we need the square of the determinant, maybe there's a way to relate that to the areas or something else?

But maybe I should think more directly. Let me write down the determinant. The determinant is:

| x1 y1 1 |
| x2 y2 1 |
| x3 y3 1 |

The determinant can be calculated as x1(y2 - y3) - y1(x2 - x3) + 1(x2y3 - x3y2). Alternatively, expanding along the third column, since it has all ones. Let me recall the formula for the determinant:

For a matrix:

| a b c |
| d e f |
| g h i |

The determinant is a(ei - fh) - b(di - fg) + c(dh - eg). So applying that here:

Determinant = x1(y2*1 - y3*1) - y1(x2*1 - x3*1) + 1(x2*y3 - x3*y2)

Simplify:

= x1(y2 - y3) - y1(x2 - x3) + (x2y3 - x3y2)

Hmm, maybe I can factor this expression. Let me see:

Let me note that x2y3 - x3y2 is the determinant of the 2x2 matrix with columns (x2, y2) and (x3, y3). Which is also twice the area of the triangle formed by the origin, (x2,y2), and (x3,y3). But maybe that's not directly helpful here.

Alternatively, maybe I can think of the determinant as related to the area of the triangle formed by the three points. Let me recall that the area of the triangle with vertices (x1,y1), (x2,y2), (x3,y3) is given by (1/2)|determinant|, where determinant is the same as above. Wait, actually, the formula for the area is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Which is similar to the determinant expression. Let me check:

Expanding the determinant:

x1(y2 - y3) - y1(x2 - x3) + (x2y3 - x3y2)

Let me rearrange terms:

= x1y2 - x1y3 - y1x2 + y1x3 + x2y3 - x3y2

Grouping terms:

= (x1y2 - x2y1) + (x2y3 - x3y2) + (x3y1 - x1y3)

Wait, but the area formula is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Let me compute that:

x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = x1y2 - x1y3 + x2y3 - x2y1 + x3y1 - x3y2, which is the same as the determinant expression above. Therefore, the determinant is equal to twice the area of the triangle. Therefore, the determinant squared would be four times the square of the area. But the problem is asking for the square of the determinant, so if I can find the area, then square it and multiply by 4? Wait, no. Wait, determinant is twice the area, so determinant squared is 4*(area)^2. Therefore, if I can find the area, then compute 4*(area)^2. But how do I find the area?

Alternatively, maybe there's a way to relate the given distances between the points to the area. The three distances given are 9, 16, 25. Let me note that 9, 16, 25 are squares of 3, 4, 5. So the triangle formed by the three points is a right triangle? Wait, if the sides are 3, 4, 5, then yes, it's a right triangle. Wait, but the problem doesn't say that the triangle is right-angled. Wait, but the distances between the points are 3, 4, 5. So if the three sides are 3, 4, 5, then it's a right triangle. Therefore, the area would be (3*4)/2 = 6. Therefore, determinant squared would be 4*(6)^2 = 4*36 = 144? But wait, is that correct?

Wait, but hold on. The problem says that the distances between the points are 9, 16, 25. So the sides are 3, 4, 5. Therefore, the triangle is right-angled. Therefore, area is 6. Therefore determinant squared is 4*(area)^2 = 4*36 = 144. Therefore, the answer is 144? But wait, is that necessarily true?

Wait, but hold on. The problem doesn't specify that the triangle is right-angled. Wait, but if the three sides are 3, 4, 5, then by Pythagoras, it must be a right triangle. Because 3² + 4² = 5². Therefore, the triangle is right-angled. Therefore, area is 6. Therefore determinant squared is 4*(6)^2 = 144. Therefore, answer is 144? But wait, but maybe the triangle isn't right-angled? Wait, but if the distances between the points are 3, 4, 5, then the triangle must be right-angled. Because the sides satisfy Pythagoras. Therefore, the area is 6. Therefore determinant squared is 4*(area)^2 = 144. Therefore, answer is 144. But wait, but maybe I need to check if the determinant is twice the area. Let me confirm.

Yes, the area of the triangle is (1/2)|determinant|. Therefore determinant is twice the area. Therefore determinant squared is 4*(area)^2. Therefore, if area is 6, determinant squared is 4*36=144. Therefore, answer is 144. But wait, but maybe the triangle isn't right-angled? Wait, but the distances are 3, 4, 5. So sides are 3, 4, 5. Therefore, it's a right triangle. Therefore, area is 6. Therefore, determinant squared is 144. Therefore, answer is 144. But let me check again.

But maybe there's a mistake here. Because the problem says that the three distances squared are 9, 16, 25. So the sides are 3, 4, 5. Therefore, the triangle is right-angled. Therefore, area is 6. Therefore determinant squared is 4*(area)^2 = 144. Therefore, answer is 144. But maybe I need to think more carefully. Because maybe the determinant isn't directly twice the area? Let me check with coordinates.

Suppose I take coordinates for a right triangle. Let me take points (0,0), (3,0), (0,4). Then the distances between them are 3, 4, 5. Let me compute the determinant:

| 0 0 1 |
| 3 0 1 |
| 0 4 1 |

Determinant is 0*(0*1 - 4*1) - 0*(3*1 - 0*1) + 1*(3*4 - 0*0) = 0 - 0 + 1*(12 - 0) = 12. Then determinant squared is 144. Which matches the previous result. The area of the triangle is (1/2)*3*4=6, so determinant is 12, which is twice the area. Therefore determinant squared is 144. Therefore, answer is 144. Therefore, maybe the answer is 144. But wait, but the problem didn't specify that the triangle is right-angled. However, given the distances between the points, it must be right-angled. Therefore, answer is 144. But maybe there's a different approach?

Alternatively, maybe I need to use coordinates. Let me try to assign coordinates to the points. Let me assume that the points are A(x1,y1), B(x2,y2), C(x3,y3). The distances AB=3, AC=4, BC=5. Wait, but AB squared is 9, AC squared is 16, BC squared is 25. Wait, but AB squared is (x1 - x2)^2 + (y1 - y2)^2 = 9, AC squared is (x1 - x3)^2 + (y1 - y3)^2 = 16, BC squared is (x2 - x3)^2 + (y2 - y3)^2 = 25. So AB=3, AC=4, BC=5. Therefore, triangle ABC has sides 3,4,5. Therefore, right-angled. Therefore, area is 6. Therefore determinant squared is 144. Therefore answer is 144. Therefore, maybe that's the answer. But maybe I need to check if there's another way to compute the determinant squared without assuming the triangle is right-angled?

Alternatively, maybe using coordinates. Let me try to compute the determinant squared. Let me denote the determinant as D. Then D^2 is what we need. Let me recall that D^2 is equal to the square of the area times 4. But if I can compute D^2 using the given distances. Alternatively, maybe using vectors.

Let me think. Let me consider vectors AB and AC. Let me denote vector AB = (x2 - x1, y2 - y1), vector AC = (x3 - x1, y3 - y1). Then the area of the triangle is (1/2)|AB × AC|, which is (1/2)| (x2 - x1)(y3 - y1) - (x3 - x1)(y2 - y1) |. Which is (1/2)| determinant |. Therefore, determinant is twice the area. Therefore, determinant squared is 4*(area)^2. Therefore, if I can compute the area, then I can find determinant squared. But how?

Alternatively, since we know the lengths of AB, AC, and BC, we can use Heron's formula. Let me try that. The sides are 3, 4, 5. The semi-perimeter s = (3 + 4 + 5)/2 = 6. Then area = sqrt[s(s - a)(s - b)(s - c)] = sqrt[6*(6 - 3)(6 - 4)(6 - 5)] = sqrt[6*3*2*1] = sqrt[36] = 6. Therefore, area is 6. Therefore determinant squared is 4*(6)^2 = 144. Therefore, answer is 144. Therefore, regardless of the coordinates, the answer is 144. Therefore, maybe that's the answer.

But wait, but the problem didn't specify that the triangle is right-angled. However, since the sides are 3, 4, 5, which is a Pythagorean triple, so it must be right-angled. Therefore, area is 6. Therefore, determinant squared is 144. Therefore, answer is 144. Therefore, maybe that's the answer. However, maybe there's a mistake here. Let me think again.

Wait, but suppose that the three points are not in the plane? Wait, no, the problem says that x1, x2, x3, y1, y2, y3 are real numbers, so they are in the plane. Therefore, the triangle is in the plane, and with sides 3, 4, 5, so right-angled. Therefore, area is 6. Therefore determinant squared is 144. Therefore, answer is 144. Therefore, maybe that's the answer.

But let me think again. Maybe there's another way to compute the determinant squared. Let me recall that the determinant squared is equal to the square of the area times 4. But also, there is a formula involving the side lengths. Let me recall that for a triangle with sides a, b, c, the area is given by Heron's formula, which we did. Alternatively, using the formula (1/4) * sqrt( (a + b + c)(-a + b + c)(a - b + c)(a + b - c) ). Which is the same as Heron's formula. Therefore, in this case, the area is 6, so determinant squared is 4*(6)^2 = 144. Therefore, answer is 144.

Alternatively, maybe I can use coordinates. Let me try to assign coordinates to the points. Let me assume that point A is at (0, 0), point B is at (3, 0), since AB = 3. Then point C is somewhere such that AC = 4 and BC = 5. Let me find coordinates of point C. Let me denote point C as (x, y). Then distance from A to C is sqrt(x² + y²) = 4, so x² + y² = 16. Distance from B to C is sqrt( (x - 3)^2 + y² ) = 5, so (x - 3)^2 + y² = 25. Subtract the first equation from the second: (x - 3)^2 + y² - x² - y² = 25 - 16 => x² -6x +9 -x² = 9 => -6x +9 = 9 => -6x = 0 => x = 0. Then from x = 0, substitute into x² + y² = 16: 0 + y² = 16 => y = ±4. Therefore, point C is at (0, 4) or (0, -4). Therefore, coordinates are A(0,0), B(3,0), C(0,4). Then determinant is:

| 0 0 1 |
| 3 0 1 |
| 0 4 1 |

Which we calculated earlier as 12. Therefore determinant squared is 144. Alternatively, if point C is (0, -4), determinant would be:

| 0 0 1 |
| 3 0 1 |
| 0 -4 1 |

Which is 0*(0*1 - (-4)*1) - 0*(3*1 - 0*1) + 1*(3*(-4) - 0*0) = 0 - 0 + (-12) = -12. Therefore determinant squared is 144. Therefore, same result. Therefore, answer is 144. Therefore, regardless of the coordinates, determinant squared is 144. Therefore, answer is 144.

But wait, but the problem didn't specify coordinates. So maybe there's a way to do this without assuming coordinates? Let me think. Let me recall that the determinant squared is equal to the square of the area times 4. And since the area can be found via Heron's formula, which gives 6, so determinant squared is 144. Therefore, answer is 144. Therefore, maybe that's the answer.

But let me check another approach. Let me think about the determinant squared. The determinant squared is equal to the square of the area times 4. Alternatively, maybe using vectors. Let me denote vectors AB and AC. Then the area is (1/2)|AB × AC|. Therefore, determinant is twice the area, so determinant squared is 4*(area)^2. Therefore, if I can compute (area)^2, then multiply by 4. But how?

Alternatively, using the formula for the area in terms of the side lengths. Which is Heron's formula. Therefore, as above, area squared is 36, so determinant squared is 4*36=144. Therefore, answer is 144. Therefore, regardless of the coordinate system, answer is 144. Therefore, answer is 144.

But maybe there's another way to compute determinant squared using the given distances. Let me think. Let me denote the determinant as D. Then D^2 is what we need. Let me recall that D is the determinant of the matrix:

[ x1 y1 1 ]

[ x2 y2 1 ]

[ x3 y3 1 ]

Which can be written as:

D = x1(y2 - y3) - y1(x2 - x3) + (x2y3 - x3y2)

Alternatively, D = (x1(y2 - y3) - y1(x2 - x3)) + (x2y3 - x3y2)

But maybe I can relate this to the given distances. Let me note that (x1 - x2)^2 + (y1 - y2)^2 = 9, which is AB^2. Similarly for others. Let me denote AB^2 = 9, AC^2 = 16, BC^2 = 25.

Alternatively, maybe I can use coordinates. Let me consider coordinates with A at (0,0), B at (a, 0), and C at (d, e). Then AB^2 = a² = 9 => a = 3. AC^2 = d² + e² = 16. BC^2 = (d - a)^2 + e² = 25. Then, as before, solving gives d = 0, e = ±4. Therefore, coordinates of C are (0, 4) or (0, -4). Then determinant is as before, which gives D = ±12, so D² = 144. Therefore, answer is 144. Therefore, same result.

Alternatively, maybe using vectors. Let me think. Let me denote vectors from point A to B as vector AB, and from A to C as vector AC. Then the area is (1/2)|AB × AC|. The cross product AB × AC is |AB||AC|sin(theta), where theta is the angle between AB and AC. Therefore, area is (1/2)|AB||AC|sin(theta). Therefore, determinant squared is 4*(area)^2 = 4*(1/4)|AB|²|AC|²sin²(theta) = |AB|²|AC|²sin²(theta). But also, from the law of cosines, BC² = AB² + AC² - 2AB*AC*cos(theta). Therefore, 25 = 9 + 16 - 2*3*4*cos(theta). Wait, 9 + 16 = 25. Therefore, 25 = 25 - 24 cos(theta). Therefore, 0 = -24 cos(theta) => cos(theta) = 0. Therefore, theta = 90 degrees. Therefore, sin(theta) = 1. Therefore, determinant squared is |AB|²|AC|²sin²(theta) = 9*16*1 = 144. Therefore, answer is 144. Therefore, this approach also gives 144. Therefore, this seems to confirm the answer.

Therefore, regardless of the method, the answer is 144. Therefore, I think the answer is 144. Therefore, I can write that as the final answer.

**Final Answer**
\boxed{144}

To solve the problem, we are given three equations involving the squared distances between three points in the plane:

$$
\begin{align*}
(x_1 - x_2)^2 + (y_1 - y_2)^2 &= 9, \\
(x_1 - x_3)^2 + (y_1 - y_3)^2 &= 16, \\
(x_2 - x_3)^2 + (y_2 - y_3)^2 &= 25.
\end{align*}
$$

These represent the squared distances between the points $ A = (x_1, y_1) $, $ B = (x_2, y_2) $, and $ C = (x_3, y_3) $, so the side lengths of triangle $ ABC $ are:

$$
AB = \sqrt{9} = 3, \quad AC = \sqrt{16} = 4, \quad BC = \sqrt{25} = 5.
$$

---

### Step 1: Recognize the Triangle Type

The side lengths $ 3, 4, 5 $ form a **right triangle**, since:

$$
3^2 + 4^2 = 9 + 16 = 25 = 5^2.
$$

Thus, triangle $ ABC $ is a **right triangle** with the right angle at point $ A $, assuming the order of the points is such that $ AB $ and $ AC $ are the legs.

---

### Step 2: Compute the Area of the Triangle

For a right triangle, the area is:

$$
\text{Area} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2 = \frac{1}{2} \times 3 \times 4 = 6.
$$

---

### Step 3: Relate the Area to the Determinant

The determinant of the matrix:

$$
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix}
$$

represents **twice the area** of the triangle formed by the points $ A, B, C $. That is:

$$
\text{Determinant} = 2 \times \text{Area} = 2 \times 6 = 12.
$$

Therefore, the square of the determinant is:

$$
\left( \text{Determinant} \right)^2 = 12^2 = 144.
$$

---

### Final Answer

$$
\boxed{144}
$$

[Assistant]
Explanation:
Okay, so I need to find the square of the determinant of a 3x3 matrix with columns (x1, y1, 1), (x2, y2, 1), and (x3, y3, 1). The determinant squared is what's being asked. The problem gives me three equations involving the distances between three points (x1,y1), (x2,y2), and (x3,y3). Specifically, the distances squared between each pair are 9, 16, and 25.

First, let me recall that the determinant of a 3x3 matrix with columns as points (x1,y1,1), etc., relates to the area of the triangle formed by those three points. Wait, actually, the determinant itself is twice the area of the triangle, right? Because the area can be calculated using the shoelace formula, and the determinant is related to that. But since we need the square of the determinant, maybe there's a way to relate that to the areas or something else?

But maybe I should think more directly. Let me write down the determinant. The determinant is:

| x1 y1 1 |
| x2 y2 1 |
| x3 y3 1 |

The determinant can be calculated as x1(y2 - y3) - y1(x2 - x3) + 1(x2y3 - x3y2). Alternatively, expanding along the third column, since it has all ones. Let me recall the formula for the determinant:

For a matrix:

| a b c |
| d e f |
| g h i |

The determinant is a(ei - fh) - b(di - fg) + c(dh - eg). So applying that here:

Determinant = x1(y2*1 - y3*1) - y1(x2*1 - x3*1) + 1(x2*y3 - x3*y2)

Simplify:

= x1(y2 - y3) - y1(x2 - x3) + (x2y3 - x3y2)

Hmm, maybe I can factor this expression. Let me see:

Let me note that x2y3 - x3y2 is the determinant of the 2x2 matrix with columns (x2, y2) and (x3, y3). Which is also twice the area of the triangle formed by the origin, (x2,y2), and (x3,y3). But maybe that's not directly helpful here.

Alternatively, maybe I can think of the determinant as related to the area of the triangle formed by the three points. Let me recall that the area of the triangle with vertices (x1,y1), (x2,y2), (x3,y3) is given by (1/2)|determinant|, where determinant is the same as above. Wait, actually, the formula for the area is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Which is similar to the determinant expression. Let me check:

Expanding the determinant:

x1(y2 - y3) - y1(x2 - x3) + (x2y3 - x3y2)

Let me rearrange terms:

= x1y2 - x1y3 - y1x2 + y1x3 + x2y3 - x3y2

Grouping terms:

= (x1y2 - x2y1) + (x2y3 - x3y2) + (x3y1 - x1y3)

Wait, but the area formula is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Let me compute that:

x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) = x1y2 - x1y3 + x2y3 - x2y1 + x3y1 - x3y2, which is the same as the determinant expression above. Therefore, the determinant is equal to twice the area of the triangle. Therefore, the determinant squared would be four times the square of the area. But the problem is asking for the square of the determinant, so if I can find the area, then square it and multiply by 4? Wait, no. Wait, determinant is twice the area, so determinant squared is 4*(area)^2. Therefore, if I can find the area, then compute 4*(area)^2. But how do I find the area?

Alternatively, maybe there's a way to relate the given distances between the points to the area. The three distances given are 9, 16, 25. Let me note that 9, 16, 25 are squares of 3, 4, 5. So the triangle formed by the three points is a right triangle? Wait, if the sides are 3, 4, 5, then yes, it's a right triangle. Wait, but the problem doesn't say that the triangle is right-angled. Wait, but the distances between the points are 3, 4, 5. So if the three sides are 3, 4, 5, then it's a right triangle. Therefore, the area would be (3*4)/2 = 6. Therefore, determinant squared would be 4*(6)^2 = 4*36 = 144? But wait, is that correct?

Wait, but hold on. The problem says that the distances between the points are 9, 16, 25. So the sides are 3, 4, 5. Therefore, the triangle is right-angled. Therefore, area is 6. Therefore determinant squared is 4*(area)^2 = 4*36 = 144. Therefore, the answer is 144? But wait, is that necessarily true?

Wait, but hold on. The problem doesn't specify that the triangle is right-angled. Wait, but if the three sides are 3, 4, 5, then by Pythagoras, it must be a right triangle. Because 3² + 4² = 5². Therefore, the triangle is right-angled. Therefore, area is 6. Therefore determinant squared is 4*(6)^2 = 144. Therefore, answer is 144? But wait, but maybe the triangle isn't right-angled? Wait, but if the distances between the points are 3, 4, 5, then the triangle must be right-angled. Because the sides satisfy Pythagoras. Therefore, the area is 6. Therefore determinant squared is 4*(area)^2 = 144. Therefore, answer is 144. But wait, but maybe I need to check if the determinant is twice the area. Let me confirm.

Yes, the area of the triangle is (1/2)|determinant|. Therefore determinant is twice the area. Therefore determinant squared is 4*(area)^2. Therefore, if area is 6, determinant squared is 4*36=144. Therefore, answer is 144. But wait, but maybe the triangle isn't right-angled? Wait, but the distances are 3, 4, 5. So sides are 3, 4, 5. Therefore, it's a right triangle. Therefore, area is 6. Therefore, determinant squared is 144. Therefore, answer is 144. But let me check again.

But maybe there's a mistake here. Because the problem says that the three distances squared are 9, 16, 25. So the sides are 3, 4, 5. Therefore, the triangle is right-angled. Therefore, area is 6. Therefore determinant squared is 4*(area)^2 = 144. Therefore, answer is 144. But maybe I need to think more carefully. Because maybe the determinant isn't directly twice the area? Let me check with coordinates.

Suppose I take coordinates for a right triangle. Let me take points (0,0), (3,0), (0,4). Then the distances between them are 3, 4, 5. Let me compute the determinant:

| 0 0 1 |
| 3 0 1 |
| 0 4 1 |

Determinant is 0*(0*1 - 4*1) - 0*(3*1 - 0*1) + 1*(3*4 - 0*0) = 0 - 0 + 1*(12 - 0) = 12. Then determinant squared is 144. Which matches the previous result. The area of the triangle is (1/2)*3*4=6, so determinant is 12, which is twice the area. Therefore determinant squared is 144. Therefore, answer is 144. Therefore, maybe the answer is 144. But wait, but the problem didn't specify that the triangle is right-angled. However, given the distances between the points, it must be right-angled. Therefore, answer is 144. But maybe there's a different approach?

Alternatively, maybe I need to use coordinates. Let me try to assign coordinates to the points. Let me assume that the points are A(x1,y1), B(x2,y2), C(x3,y3). The distances AB=3, AC=4, BC=5. Wait, but AB squared is 9, AC squared is 16, BC squared is 25. Wait, but AB squared is (x1 - x2)^2 + (y1 - y2)^2 = 9, AC squared is (x1 - x3)^2 + (y1 - y3)^2 = 16, BC squared is (x2 - x3)^2 + (y2 - y3)^2 = 25. So AB=3, AC=4, BC=5. Therefore, triangle ABC has sides 3,4,5. Therefore, right-angled. Therefore, area is 6. Therefore determinant squared is 144. Therefore answer is 144. Therefore, maybe that's the answer. But maybe I need to check if there's another way to compute the determinant squared without assuming the triangle is right-angled?

Alternatively, maybe using coordinates. Let me try to compute the determinant squared. Let me denote the determinant as D. Then D^2 is what we need. Let me recall that D^2 is equal to the square of the area times 4. But if I can compute D^2 using the given distances. Alternatively, maybe using vectors.

Let me think. Let me consider vectors AB and AC. Let me denote vector AB = (x2 - x1, y2 - y1), vector AC = (x3 - x1, y3 - y1). Then the area of the triangle is (1/2)|AB × AC|, which is (1/2)| (x2 - x1)(y3 - y1) - (x3 - x1)(y2 - y1) |. Which is (1/2)| determinant |. Therefore, determinant is twice the area. Therefore, determinant squared is 4*(area)^2. Therefore, if I can compute the area, then I can find determinant squared. But how?

Alternatively, since we know the lengths of AB, AC, and BC, we can use Heron's formula. Let me try that. The sides are 3, 4, 5. The semi-perimeter s = (3 + 4 + 5)/2 = 6. Then area = sqrt[s(s - a)(s - b)(s - c)] = sqrt[6*(6 - 3)(6 - 4)(6 - 5)] = sqrt[6*3*2*1] = sqrt[36] = 6. Therefore, area is 6. Therefore determinant squared is 4*(6)^2 = 144. Therefore, answer is 144. Therefore, regardless of the coordinates, the answer is 144. Therefore, maybe that's the answer.

But wait, but the problem didn't specify that the triangle is right-angled. However, since the sides are 3, 4, 5, which is a Pythagorean triple, so it must be right-angled. Therefore, area is 6. Therefore, determinant squared is 144. Therefore, answer is 144. Therefore, maybe that's the answer. However, maybe there's a mistake here. Let me think again.

Wait, but suppose that the three points are not in the plane? Wait, no, the problem says that x1, x2, x3, y1, y2, y3 are real numbers, so they are in the plane. Therefore, the triangle is in the plane, and with sides 3, 4, 5, so right-angled. Therefore, area is 6. Therefore determinant squared is 144. Therefore, answer is 144. Therefore, maybe that's the answer.

But let me think again. Maybe there's another way to compute the determinant squared. Let me recall that the determinant squared is equal to the square of the area times 4. But also, there is a formula involving the side lengths. Let me recall that for a triangle with sides a, b, c, the area is given by Heron's formula, which we did. Alternatively, using the formula (1/4) * sqrt( (a + b + c)(-a + b + c)(a - b + c)(a + b - c) ). Which is the same as Heron's formula. Therefore, in this case, the area is 6, so determinant squared is 4*(6)^2 = 144. Therefore, answer is 144.

Alternatively, maybe I can use coordinates. Let me try to assign coordinates to the points. Let me assume that point A is at (0, 0), point B is at (3, 0), since AB = 3. Then point C is somewhere such that AC = 4 and BC = 5. Let me find coordinates of point C. Let me denote point C as (x, y). Then distance from A to C is sqrt(x² + y²) = 4, so x² + y² = 16. Distance from B to C is sqrt( (x - 3)^2 + y² ) = 5, so (x - 3)^2 + y² = 25. Subtract the first equation from the second: (x - 3)^2 + y² - x² - y² = 25 - 16 => x² -6x +9 -x² = 9 => -6x +9 = 9 => -6x = 0 => x = 0. Then from x = 0, substitute into x² + y² = 16: 0 + y² = 16 => y = ±4. Therefore, point C is at (0, 4) or (0, -4). Therefore, coordinates are A(0,0), B(3,0), C(0,4). Then determinant is:

| 0 0 1 |
| 3 0 1 |
| 0 4 1 |

Which we calculated earlier as 12. Therefore determinant squared is 144. Alternatively, if point C is (0, -4), determinant would be:

| 0 0 1 |
| 3 0 1 |
| 0 -4 1 |

Which is 0*(0*1 - (-4)*1) - 0*(3*1 - 0*1) + 1*(3*(-4) - 0*0) = 0 - 0 + (-12) = -12. Therefore determinant squared is 144. Therefore, same result. Therefore, answer is 144. Therefore, regardless of the coordinates, determinant squared is 144. Therefore, answer is 144.

But wait, but the problem didn't specify coordinates. So maybe there's a way to do this without assuming coordinates? Let me think. Let me recall that the determinant squared is equal to the square of the area times 4. And since the area can be found via Heron's formula, which gives 6, so determinant squared is 144. Therefore, answer is 144. Therefore, maybe that's the answer.

But let me check another approach. Let me think about the determinant squared. The determinant squared is equal to the square of the area times 4. Alternatively, maybe using vectors. Let me denote vectors AB and AC. Then the area is (1/2)|AB × AC|. Therefore, determinant is twice the area, so determinant squared is 4*(area)^2. Therefore, if I can compute (area)^2, then multiply by 4. But how?

Alternatively, using the formula for the area in terms of the side lengths. Which is Heron's formula. Therefore, as above, area squared is 36, so determinant squared is 4*36=144. Therefore, answer is 144. Therefore, regardless of the coordinate system, answer is 144. Therefore, answer is 144.

But maybe there's another way to compute determinant squared using the given distances. Let me think. Let me denote the determinant as D. Then D^2 is what we need. Let me recall that D is the determinant of the matrix:

[ x1 y1 1 ]

[ x2 y2 1 ]

[ x3 y3 1 ]

Which can be written as:

D = x1(y2 - y3) - y1(x2 - x3) + (x2y3 - x3y2)

Alternatively, D = (x1(y2 - y3) - y1(x2 - x3)) + (x2y3 - x3y2)

But maybe I can relate this to the given distances. Let me note that (x1 - x2)^2 + (y1 - y2)^2 = 9, which is AB^2. Similarly for others. Let me denote AB^2 = 9, AC^2 = 16, BC^2 = 25.

Alternatively, maybe I can use coordinates. Let me consider coordinates with A at (0,0), B at (a, 0), and C at (d, e). Then AB^2 = a² = 9 => a = 3. AC^2 = d² + e² = 16. BC^2 = (d - a)^2 + e² = 25. Then, as before, solving gives d = 0, e = ±4. Therefore, coordinates of C are (0, 4) or (0, -4). Then determinant is as before, which gives D = ±12, so D² = 144. Therefore, answer is 144. Therefore, same result.

Alternatively, maybe using vectors. Let me think. Let me denote vectors from point A to B as vector AB, and from A to C as vector AC. Then the area is (1/2)|AB × AC|. The cross product AB × AC is |AB||AC|sin(theta), where theta is the angle between AB and AC. Therefore, area is (1/2)|AB||AC|sin(theta). Therefore, determinant squared is 4*(area)^2 = 4*(1/4)|AB|²|AC|²sin²(theta) = |AB|²|AC|²sin²(theta). But also, from the law of cosines, BC² = AB² + AC² - 2AB*AC*cos(theta). Therefore, 25 = 9 + 16 - 2*3*4*cos(theta). Wait, 9 + 16 = 25. Therefore, 25 = 25 - 24 cos(theta). Therefore, 0 = -24 cos(theta) => cos(theta) = 0. Therefore, theta = 90 degrees. Therefore, sin(theta) = 1. Therefore, determinant squared is |AB|²|AC|²sin²(theta) = 9*16*1 = 144. Therefore, answer is 144. Therefore, this approach also gives 144. Therefore, this seems to confirm the answer.

Therefore, regardless of the method, the answer is 144. Therefore, I think the answer is 144. Therefore, I can write that as the final answer.

**Final Answer**
\boxed{144}

To solve the problem, we are given three equations involving the squared distances between three points in the plane:

$$
\begin{align*}
(x_1 - x_2)^2 + (y_1 - y_2)^2 &= 9, \\
(x_1 - x_3)^2 + (y_1 - y_3)^2 &= 16, \\
(x_2 - x_3)^2 + (y_2 - y_3)^2 &= 25.
\end{align*}
$$

These represent the squared distances between the points $ A = (x_1, y_1) $, $ B = (x_2, y_2) $, and $ C = (x_3, y_3) $, so the side lengths of triangle $ ABC $ are:

$$
AB = \sqrt{9} = 3, \quad AC = \sqrt{16} = 4, \quad BC = \sqrt{25} = 5.
$$

---

### Step 1: Recognize the Triangle Type

The side lengths $ 3, 4, 5 $ form a **right triangle**, since:

$$
3^2 + 4^2 = 9 + 16 = 25 = 5^2.
$$

Thus, triangle $ ABC $ is a **right triangle** with the right angle at point $ A $, assuming the order of the points is such that $ AB $ and $ AC $ are the legs.

---

### Step 2: Compute the Area of the Triangle

For a right triangle, the area is:

$$
\text{Area} = \frac{1}{2} \times \text{leg}_1 \times \text{leg}_2 = \frac{1}{2} \times 3 \times 4 = 6.
$$

---

### Step 3: Relate the Area to the Determinant

The determinant of the matrix:

$$
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix}
$$

represents **twice the area** of the triangle formed by the points $ A, B, C $. That is:

$$
\text{Determinant} = 2 \times \text{Area} = 2 \times 6 = 12.
$$

Therefore, the square of the determinant is:

$$
\left( \text{Determinant} \right)^2 = 12^2 = 144.
$$

---

### Final Answer

$$
\boxed{144}
$$

Solution:
In general,
\[\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\]is the signed area of the triangle with vertices at $(x_1,y_1),$ $(x_2,y_2),$ and $(x_3,y_3).$  (The area is signed, i.e. it can be positive or negative, depending on the orientation of the triangle.)  Here, the sides of the triangle are 3, 4, and 5, which is a right triangle.  Therefore, its area is $\frac{1}{2} \cdot 3 \cdot 4 = 6.$  Then
\[\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} = \pm 12,\]so
\[\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}^2 = \boxed{144}.\]


Explanation:
Okay, so I need to find the area of triangle ENG in triangle ABC, where the medians AD, BE, and CF intersect at centroid G. There's a line through G parallel to BC that intersects AB and AC at M and N, respectively. The area of triangle ABC is given as 144. Hmm, let me try to visualize this first.

First, let me recall that the centroid divides each median into a ratio of 2:1, with the longer part being closer to the vertex. So, for example, AG:GD = 2:1, BG:GE = 2:1, and CG:GF = 2:1. Also, the line through G parallel to BC... since BC is the base, maybe this line MN is somewhere inside the triangle, cutting AB and AC.

Let me try to sketch this mentally. Triangle ABC, with centroid G. Drawing a line through G parallel to BC, which would be a line that's parallel to the base BC, so it should be somewhere between the base and the vertex A. Since it's parallel to BC, maybe it's similar to triangle ABC? But scaled down?

But the problem mentions points M and N on AB and AC. So MN is parallel to BC, passing through G. So maybe I can use properties of similar triangles or coordinate geometry here?

Alternatively, maybe mass point geometry? But since the area is given, coordinate geometry might be more straightforward. Let me try coordinate geometry.

Let me assign coordinates to triangle ABC. Let me place point A at (0, 0), B at (2b, 0), and C at (2c, 2d). Wait, but maybe it's better to use coordinates that make calculations easier. Since medians are involved, maybe using coordinates where centroid is at (0, 0)? Wait, but maybe not. Let me think.

Alternatively, let me take coordinates such that BC is horizontal? Wait, but since the line MN is parallel to BC, maybe it's easier to have BC horizontal. Let me try that.

Let me set coordinate system with point B at (0, 0), point C at (c, 0), so BC is along the x-axis. Then point A can be at some coordinates (a, b). Then centroid G is the average of the coordinates of A, B, C. So coordinates of G would be ((a + 0 + c)/3, (b + 0 + 0)/3) = ((a + c)/3, b/3).

Now, the line through G parallel to BC. Since BC is horizontal, the line through G parallel to BC is also horizontal. Therefore, it has equation y = b/3. This line intersects AB and AC at points M and N. Let me find coordinates of M and N.

First, find equations of AB and AC.

Equation of AB: connects A(a, b) and B(0, 0). The parametric equations can be written as x = a*t, y = b*t, where t ranges from 0 to 1.

Similarly, equation of AC: connects A(a, b) and C(c, 0). Parametric equations can be written as x = a + (c - a)*s, y = b - b*s, where s ranges from 0 to 1.

But maybe it's easier to find the intersection points with the line y = b/3.

For AB: Let me find where y = b/3. On AB, the line from (0, 0) to (a, b). The slope of AB is (b - 0)/(a - 0) = b/a. So equation is y = (b/a)x. Setting y = b/3, we have b/3 = (b/a)x => x = a/3. Therefore, point M is at (a/3, b/3).

Similarly, for AC: The line from A(a, b) to C(c, 0). The slope is (0 - b)/(c - a) = -b/(c - a). Equation of AC: y - b = [-b/(c - a)](x - a). Let me solve for y when y = b/3.

So, b/3 - b = [-b/(c - a)](x - a)

=> -2b/3 = [-b/(c - a)](x - a)

Multiply both sides by (c - a)/(-b):

(-2b/3) * (c - a)/(-b) = x - a

Simplify:

(2/3)(c - a) = x - a

Therefore, x = a + (2/3)(c - a) = (a/3) + (2c)/3 = (a + 2c)/3

Therefore, point N is at ((a + 2c)/3, b/3)

So coordinates of M: (a/3, b/3), coordinates of N: ((a + 2c)/3, b/3). Therefore, MN is the line segment between these two points, which is horizontal at y = b/3.

Now, the problem mentions triangle ENG. Wait, but where is point E? Wait, the medians are AD, BE, and CF. So BE is a median from B to AC. So point E is the midpoint of AC. Wait, yes. Since BE is a median, E is the midpoint of AC.

Similarly, D is the midpoint of BC, and F is the midpoint of AB.

So coordinates of E: midpoint of AC. Coordinates of A(a, b) and C(c, 0), so midpoint E is ((a + c)/2, (b + 0)/2) = ((a + c)/2, b/2).

Wait, so point E is ((a + c)/2, b/2). Then point G is ((a + c)/3, b/3). So we need to find the area of triangle ENG, where N is ((a + 2c)/3, b/3), E is ((a + c)/2, b/2), and G is ((a + c)/3, b/3).

Wait, so triangle ENG has coordinates:

E: ((a + c)/2, b/2)

N: ((a + 2c)/3, b/3)

G: ((a + c)/3, b/3)

So to find the area of triangle ENG, we can use coordinates. Let me recall the formula for area given three coordinates.

The area can be calculated using determinant formula:

Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Let me assign:

E: (x1, y1) = ((a + c)/2, b/2)

N: (x2, y2) = ((a + 2c)/3, b/3)

G: (x3, y3) = ((a + c)/3, b/3)

So plug into the formula:

Area = (1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

Compute each term:

First term: x1(y2 - y3) = [(a + c)/2] * [ (b/3 - b/3) ] = [(a + c)/2] * 0 = 0

Second term: x2(y3 - y1) = [(a + 2c)/3] * [ (b/3 - b/2) ] = [(a + 2c)/3] * [ (-b/6) ] = -b(a + 2c)/18

Third term: x3(y1 - y2) = [(a + c)/3] * [ (b/2 - b/3) ] = [(a + c)/3] * [ b/6 ] = b(a + c)/18

Therefore, adding them up:

0 + (-b(a + 2c)/18) + (b(a + c)/18) = [ -b(a + 2c) + b(a + c) ] / 18 = [ -ab - 2bc + ab + bc ] / 18 = (-bc)/18

Taking absolute value and multiplying by 1/2:

Area = (1/2) * | -bc / 18 | = (1/2)(bc/18) = bc/36

So the area of triangle ENG is bc/36. Now, we need to relate this to the area of triangle ABC, which is given as 144.

What is the area of triangle ABC? Using coordinates, with A(a, b), B(0, 0), C(c, 0). The area can be calculated as (base * height)/2. The base BC is from (0, 0) to (c, 0), so length c. The height is the y-coordinate of A, which is b. Wait, but if A is at (a, b), then the base BC is along x-axis from (0, 0) to (c, 0), but point A is at (a, b). Wait, but the area of triangle ABC is |(Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By))/2|.

Plugging in coordinates:

Area = (1/2)| a*(0 - 0) + 0*(0 - b) + c*(b - 0) | = (1/2)| 0 + 0 + bc | = (1/2)| bc | = bc/2

Given that the area is 144, so bc/2 = 144 => bc = 288

Therefore, area of triangle ENG is bc/36 = 288 / 36 = 8

Wait, so the area is 8? Let me check my steps again to make sure.

First, coordinates assigned: A(a, b), B(0, 0), C(c, 0). Then centroid G is ((a + c)/3, b/3). Then line through G parallel to BC (which is horizontal) is y = b/3. Found points M and N on AB and AC. Calculated coordinates of M as (a/3, b/3) and N as ((a + 2c)/3, b/3). Then point E is midpoint of AC: ((a + c)/2, b/2). Then coordinates of E, N, G. Then using determinant formula, found area of ENG as bc/36. Then area of ABC is bc/2 = 144 => bc = 288. Therefore, bc/36 = 8. So area of ENG is 8. That seems straightforward. But let me check if there's a mistake in coordinates.

Wait, but in my coordinate system, point A is (a, b), but if I set coordinates with B at (0, 0) and C at (c, 0), then AB is from (0, 0) to (a, b). However, if I want to make sure that the coordinates are general, maybe I need to check if there are any constraints on a? For example, if I take AB from (0, 0) to (a, b), then AC is from (a, b) to (c, 0). But maybe there's a better coordinate system where calculations are simpler?

Alternatively, maybe I can assign coordinates such that BC is of length 2 units or something? Wait, but since the area is given as 144, maybe it's better to assign coordinates such that calculations are easier. Let me try another approach.

Alternatively, let me take specific coordinates for triangle ABC. Let me assume coordinates where BC is horizontal, and centroid is at (0, 0). Wait, but maybe that complicates things. Alternatively, take coordinates where A is at (0, 0), B at (2, 0), C at (0, 2). Wait, but then area would be 2, but given area is 144. Maybe scaling later?

Wait, but maybe using vectors or ratios?

Alternatively, since the area of ABC is 144, and we found that area of ENG is bc/36, and bc/2 = 144, so bc = 288, so bc/36 = 8. Therefore, answer is 8? But let me check with specific coordinates.

Let me take specific values. Let me choose coordinates where A is (0, 6), B is (0, 0), C is (6, 0). Then area of ABC is (base * height)/2 = (6 * 6)/2 = 18. Wait, but I need area 144. Let me scale up. Let me take A(0, 12), B(0, 0), C(12, 0). Then area is (12 * 12)/2 = 72. Still not 144. Let me take A(0, 24), B(0, 0), C(12, 0). Then area is (12 * 24)/2 = 144. Perfect. So let me use this coordinate system.

So A(0, 24), B(0, 0), C(12, 0). Then centroid G is average of coordinates: ((0 + 0 + 12)/3, (24 + 0 + 0)/3) = (4, 8).

Now, line through G parallel to BC. Since BC is horizontal (from (0,0) to (12, 0)), so line through G(4, 8) parallel to BC is horizontal line y = 8. This line intersects AB and AC at M and N.

First, find equation of AB: from (0, 24) to (0, 0). Wait, AB is vertical line x = 0? Wait, but if A is (0, 24) and B is (0, 0), then AB is vertical. Then line y = 8 intersects AB at (0, 8). Wait, but the problem says the line through G parallel to BC intersects AB and AC at M and N. But if AB is vertical, then the line y = 8 intersects AB at (0, 8), which is point M. Then where does it intersect AC?

Equation of AC: from A(0, 24) to C(12, 0). The slope is (0 - 24)/(12 - 0) = -2. Equation is y - 24 = -2x => y = -2x + 24. Find intersection with y = 8:

8 = -2x + 24 => -2x = -16 => x = 8. Therefore, point N is (8, 8). Therefore, points M(0, 8) and N(8, 8).

Now, point E is the midpoint of AC. Coordinates of A(0, 24) and C(12, 0). Midpoint E is ((0 + 12)/2, (24 + 0)/2) = (6, 12).

So triangle ENG has points E(6, 12), N(8, 8), G(4, 8). Let me compute its area.

Using coordinates:

E(6, 12), N(8, 8), G(4, 8).

Using the determinant formula:

Area = (1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

Plugging in:

x1 = 6, y1 = 12

x2 = 8, y2 = 8

x3 = 4, y3 = 8

Compute:

6*(8 - 8) + 8*(8 - 12) + 4*(12 - 8)

= 6*0 + 8*(-4) + 4*(4)

= 0 - 32 + 16 = -16

Take absolute value and multiply by 1/2: (1/2)*| -16 | = 8.

So area is 8. Which matches the previous result. Since in this coordinate system, area of ABC is 144, and area of ENG is 8. Therefore, the answer is 8.

But let me check with another coordinate system to be sure. Let me take another triangle. Let me take A(0, 0), B(6, 0), C(0, 8). Then area of ABC is (6*8)/2 = 24. Let me compute area of ENG in this case, then see if the ratio holds.

First, centroid G is ((0 + 6 + 0)/3, (0 + 0 + 8)/3) = (2, 8/3).

Line through G parallel to BC. BC is from (6, 0) to (0, 8). Wait, BC is not horizontal here. Wait, in this coordinate system, BC is from B(6, 0) to C(0, 8). So slope of BC is (8 - 0)/(0 - 6) = -4/3. Therefore, line through G parallel to BC would have slope -4/3.

But the problem says the line through G parallel to BC intersects AB and AC at M and N. Wait, but in this coordinate system, AB is from A(0, 0) to B(6, 0), which is horizontal. AC is from A(0, 0) to C(0, 8), vertical. So line through G(2, 8/3) with slope -4/3. Let me find where this line intersects AB and AC.

First, equation of the line: y - 8/3 = (-4/3)(x - 2)

Simplify: y = (-4/3)x + 8/3 + 8/3 = (-4/3)x + 16/3

Intersection with AB: AB is y = 0. So set y = 0:

0 = (-4/3)x + 16/3 => (-4/3)x = -16/3 => x = 4. Therefore, point M is (4, 0).

Intersection with AC: AC is x = 0. So plug x = 0 into line equation:

y = (-4/3)(0) + 16/3 = 16/3. Therefore, point N is (0, 16/3).

Now, point E is the midpoint of AC. AC is from (0, 0) to (0, 8), so midpoint E is (0, 4).

So triangle ENG has points E(0, 4), N(0, 16/3), G(2, 8/3). Let me compute its area.

Using coordinates:

E(0, 4), N(0, 16/3), G(2, 8/3)

Using determinant formula:

Area = (1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

Plugging in:

x1 = 0, y1 = 4

x2 = 0, y2 = 16/3

x3 = 2, y3 = 8/3

Compute:

0*(16/3 - 8/3) + 0*(8/3 - 4) + 2*(4 - 16/3)

= 0 + 0 + 2*(12/3 - 16/3) = 2*(-4/3) = -8/3

Absolute value times 1/2: (1/2)*(8/3) = 4/3

Now, area of ABC is 24. What is bc/36 in this case? Wait, in my previous coordinate system, I had bc/36. Wait, but in this coordinate system, how does that relate?

Wait, in my first general coordinate system, I had A(a, b), B(0, 0), C(c, 0). Then area of ABC was bc/2. In this case, A(0, 0), B(6, 0), C(0, 8). So here, if I consider similar coordinates, but A is at (0, 0), B at (c, 0), C at (0, b). Then area is (c*b)/2. In this case, c = 6, b = 8, so area is 24, which matches. Then bc/36 would be (6*8)/36 = 48/36 = 4/3, which matches the area of triangle ENG here. Therefore, in this case, area of ENG is 4/3, and area of ABC is 24, so ratio is (4/3)/24 = 1/18. Therefore, in general, area of ENG is (area ABC)/18. Since in the first case, area ABC was 144, then area ENG is 144 / 18 = 8. Which matches. Therefore, the answer is 8.

Therefore, regardless of the coordinate system, the area of triangle ENG is 1/18 of the area of ABC. Therefore, since area ABC is 144, area ENG is 144 / 18 = 8.

But let me think again why the ratio is 1/18. In the first coordinate system, bc/36 divided by bc/2 is (bc/36)/(bc/2) = 2/36 = 1/18. So yes, the ratio is 1/18. Therefore, regardless of the triangle, the area of ENG is 1/18 of ABC. Therefore, answer is 144 / 18 = 8.

Alternatively, maybe there's a way to see this without coordinates? Let me think.

Since MN is parallel to BC and passes through centroid G. Since centroid divides the median in ratio 2:1, then the distance from G to BC is 1/3 the height of the triangle. Wait, if we consider the line MN parallel to BC through G, then the distance between MN and BC is 1/3 the height? Wait, no. Wait, the centroid is located at 1/3 of the height from the base. Wait, if we consider the height from A to BC, then centroid is located 1/3 of the height from BC. Therefore, the line MN is located at 1/3 the height from BC, so the distance between MN and BC is 1/3 the height. Therefore, the ratio of areas between triangle AMN and ABC would be (2/3)^2 = 4/9? Wait, no. Wait, if a line is drawn parallel to the base at a fraction k of the height, then the ratio of areas is k^2. But here, MN is located at 1/3 the height from BC, so the distance from A to MN is 2/3 the height. Therefore, the ratio of areas of AMN to ABC is (2/3)^2 = 4/9. Therefore, area of AMN is 4/9 * 144 = 64. Then area of the trapezoid MNBC is 144 - 64 = 80. But how does this relate to triangle ENG?

Alternatively, maybe considering coordinates again. Since MN is parallel to BC and passes through G, which divides the medians in ratio 2:1. Maybe using mass point or area ratios.

Alternatively, since E is the midpoint of AC, and N is a point on AC. Wait, in the coordinates earlier, N was located at ((a + 2c)/3, b/3). So in terms of AC, which goes from A(a, b) to C(c, 0). The point N divides AC in what ratio? Let me compute.

Parametrize AC from A to C. Let me use parameter t. Coordinates of N are ((a + 2c)/3, b/3). So starting at A(a, b), moving towards C(c, 0). The x-coordinate goes from a to c, so the difference is c - a. The x-coordinate of N is a + (c - a)*(2/3). Therefore, it's 2/3 of the way from A to C? Wait, no. Wait, if t is the fraction from A to C, then x = a + t(c - a). So solving for t:

a + t(c - a) = (a + 2c)/3

=> t(c - a) = (a + 2c)/3 - a = (a + 2c - 3a)/3 = (-2a + 2c)/3 = 2(c - a)/3

Therefore, t = [2(c - a)/3] / (c - a) ) = 2/3. Therefore, point N divides AC in ratio AN:NC = 2:1. Similarly, point M divides AB in ratio AM:MB = 2:1? Wait, in coordinates earlier, point M was at (a/3, b/3). If AB is from A(a, b) to B(0, 0). So parametrize AB with parameter t, x = a - a*t, y = b - b*t. Wait, but earlier we found that when y = b/3, t = 1/3. Therefore, point M divides AB in ratio AM:MB = 1:2. Wait, because t = 1/3 from A to B? Wait, if t is 1/3, then AM is 1/3 of AB, so MB is 2/3 of AB. Therefore, AM:MB = 1:2. Similarly, AN:NC = 2:1.

Therefore, MN is a line connecting points dividing AB and AC in ratios 1:2 and 2:1 respectively. Therefore, MN is a line segment inside the triangle.

Now, triangle ENG. Let me think about the positions of E, N, G. E is midpoint of AC, N divides AC in ratio 2:1, G is centroid.

Alternatively, since we know coordinates, maybe using vectors?

Let me consider vectors. Let me take A as origin. Let me denote vectors AB as vector b, and AC as vector c. Then coordinates:

- Point A: 0

- Point B: b

- Point C: c

- Centroid G: (A + B + C)/3 = (0 + b + c)/3 = (b + c)/3

- Point E is midpoint of AC: (A + C)/2 = (0 + c)/2 = c/2

- Point N is on AC such that AN:NC = 2:1. Therefore, vector AN = (2/3)AC = (2/3)c. Therefore, position vector of N is A + AN = 0 + (2/3)c = (2/3)c

- Point M is on AB such that AM:MB = 1:2. Therefore, vector AM = (1/3)AB = (1/3)b. Therefore, position vector of M is (1/3)b

Now, triangle ENG has points E, N, G. Let me find vectors for these points:

- E: c/2

- N: (2/3)c

- G: (b + c)/3

So vectors from E to N and E to G?

Wait, to compute area of triangle ENG, we can use cross product. In vector terms, area is (1/2)| (N - E) × (G - E) |.

Compute N - E = (2/3)c - (1/2)c = (4/6 - 3/6)c = (1/6)c

Compute G - E = (b + c)/3 - c/2 = (2(b + c) - 3c)/6 = (2b + 2c - 3c)/6 = (2b - c)/6

Then cross product (N - E) × (G - E) = (1/6)c × (2b - c)/6 = (1/36)(c × 2b - c × c) = (1/36)(2c × b - 0) = (2/36)(c × b) = (1/18)(c × b)

The magnitude of this cross product is (1/18)|c × b|. The area of triangle ENG is half of that, so (1/36)|c × b|.

But the area of triangle ABC is (1/2)|b × c|. Therefore, area of ENG is (1/36)|c × b| = (1/18)*(1/2)|b × c| = (1/18)*Area ABC. Therefore, area of ENG is 1/18 of area ABC. Since area ABC is 144, area ENG is 144 / 18 = 8.

Therefore, regardless of coordinate system, the area is 8.

This confirms the previous results. Therefore, the answer is 8.

But just to make sure, let me think if there's another way someone might approach this problem and maybe get confused. For example, someone might think that since MN is parallel to BC and passes through centroid, then maybe triangle AMN has area 4/9 of ABC, but then how does ENG relate? But in our case, ENG is a smaller triangle inside AMN or something else. But through coordinates and vectors, we saw that it's 1/18.

Alternatively, using area ratios. Since E is midpoint of AC, and N divides AC in ratio 2:1, so EN is a segment from midpoint to a point 2/3 along AC. Then, maybe considering coordinates or mass point. But since we already have multiple methods leading to 8, I think it's safe.

Therefore, the area of triangle ENG is 8.

**Final Answer**
\boxed{8}

To solve the problem, we are given a triangle $ ABC $ with medians $ AD $, $ BE $, and $ CF $ intersecting at the centroid $ G $. A line through $ G $, parallel to $ BC $, intersects $ AB $ and $ AC $ at points $ M $ and $ N $, respectively. The area of triangle $ ABC $ is 144, and we are to find the area of triangle $ ENG $, where $ E $ is the midpoint of $ AC $.

---

### Step 1: Understanding the Geometry

- The **centroid** $ G $ divides each median in the ratio $ 2:1 $, with the longer segment closer to the vertex.
- The line through $ G $ parallel to $ BC $ intersects $ AB $ at $ M $ and $ AC $ at $ N $.
- Since the line is parallel to $ BC $, triangle $ AMN $ is similar to triangle $ ABC $, and the line passes through the centroid, which lies $ \frac{1}{3} $ of the way from the base $ BC $ to the vertex $ A $.
- Therefore, the height of triangle $ AMN $ is $ \frac{2}{3} $ of the height of triangle $ ABC $, and the area of triangle $ AMN $ is $ \left(\frac{2}{3}\right)^2 = \frac{4}{9} $ of the area of triangle $ ABC $.

---

### Step 2: Coordinates and Area Calculation

To compute the area of triangle $ ENG $, we can use coordinate geometry. Let’s place triangle $ ABC $ in the coordinate plane:

- Let $ A = (0, b) $, $ B = (0, 0) $, and $ C = (c, 0) $.
- Then the area of triangle $ ABC $ is:
  $$
  \text{Area}_{ABC} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot c \cdot b = \frac{bc}{2}
  $$
  Given that this area is 144, we have:
  $$
  \frac{bc}{2} = 144 \Rightarrow bc = 288
  $$

Now, we find the coordinates of the relevant points:

- **Centroid $ G $**:
  $$
  G = \left( \frac{0 + 0 + c}{3}, \frac{b + 0 + 0}{3} \right) = \left( \frac{c}{3}, \frac{b}{3} \right)
  $$

- **Point $ E $**, the midpoint of $ AC $:
  $$
  E = \left( \frac{0 + c}{2}, \frac{b + 0}{2} \right) = \left( \frac{c}{2}, \frac{b}{2} \right)
  $$

- **Point $ N $**, the intersection of the line through $ G $ parallel to $ BC $ with $ AC $:
  Since the line is horizontal at $ y = \frac{b}{3} $, and $ AC $ has slope $ -\frac{b}{c} $, we find the x-coordinate of $ N $ by solving:
  $$
  y = -\frac{b}{c}x + b = \frac{b}{3} \Rightarrow x = \frac{2c}{3}
  $$
  So, $ N = \left( \frac{2c}{3}, \frac{b}{3} \right) $

Now, we have the coordinates of triangle $ ENG $:

- $ E = \left( \frac{c}{2}, \frac{b}{2} \right) $
- $ N = \left( \frac{2c}{3}, \frac{b}{3} \right) $
- $ G = \left( \frac{c}{3}, \frac{b}{3} \right) $

---

### Step 3: Area of Triangle $ ENG $

Using the determinant formula for the area of a triangle with vertices $ (x_1, y_1) $, $ (x_2, y_2) $, $ (x_3, y_3) $:

$$
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
$$

Substitute the coordinates:

- $ x_1 = \frac{c}{2}, y_1 = \frac{b}{2} $
- $ x_2 = \frac{2c}{3}, y_2 = \frac{b}{3} $
- $ x_3 = \frac{c}{3}, y_3 = \frac{b}{3} $

Compute each term:

- $ x_1(y_2 - y_3) = \frac{c}{2} \cdot \left( \frac{b}{3} - \frac{b}{3} \right) = 0 $
- $ x_2(y_3 - y_1) = \frac{2c}{3} \cdot \left( \frac{b}{3} - \frac{b}{2} \right) = \frac{2c}{3} \cdot \left( -\frac{b}{6} \right) = -\frac{bc}{9} $
- $ x_3(y_1 - y_2) = \frac{c}{3} \cdot \left( \frac{b}{2} - \frac{b}{3} \right) = \frac{c}{3} \cdot \frac{b}{6} = \frac{bc}{18} $

Add them:

$$
\text{Area} = \frac{1}{2} \left| 0 - \frac{bc}{9} + \frac{bc}{18} \right| = \frac{1}{2} \left| -\frac{bc}{18} \right| = \frac{1}{2} \cdot \frac{bc}{18} = \frac{bc}{36}
$$

We already know $ bc = 288 $, so:

$$
\text{Area}_{ENG} = \frac{288}{36} = 8
$$

---

### Final Answer

$$
\boxed{8}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of triangle ENG in triangle ABC, where the medians AD, BE, and CF intersect at centroid G. There's a line through G parallel to BC that intersects AB and AC at M and N, respectively. The area of triangle ABC is given as 144. Hmm, let me try to visualize this first.

First, let me recall that the centroid divides each median into a ratio of 2:1, with the longer part being closer to the vertex. So, for example, AG:GD = 2:1, BG:GE = 2:1, and CG:GF = 2:1. Also, the line through G parallel to BC... since BC is the base, maybe this line MN is somewhere inside the triangle, cutting AB and AC.

Let me try to sketch this mentally. Triangle ABC, with centroid G. Drawing a line through G parallel to BC, which would be a line that's parallel to the base BC, so it should be somewhere between the base and the vertex A. Since it's parallel to BC, maybe it's similar to triangle ABC? But scaled down?

But the problem mentions points M and N on AB and AC. So MN is parallel to BC, passing through G. So maybe I can use properties of similar triangles or coordinate geometry here?

Alternatively, maybe mass point geometry? But since the area is given, coordinate geometry might be more straightforward. Let me try coordinate geometry.

Let me assign coordinates to triangle ABC. Let me place point A at (0, 0), B at (2b, 0), and C at (2c, 2d). Wait, but maybe it's better to use coordinates that make calculations easier. Since medians are involved, maybe using coordinates where centroid is at (0, 0)? Wait, but maybe not. Let me think.

Alternatively, let me take coordinates such that BC is horizontal? Wait, but since the line MN is parallel to BC, maybe it's easier to have BC horizontal. Let me try that.

Let me set coordinate system with point B at (0, 0), point C at (c, 0), so BC is along the x-axis. Then point A can be at some coordinates (a, b). Then centroid G is the average of the coordinates of A, B, C. So coordinates of G would be ((a + 0 + c)/3, (b + 0 + 0)/3) = ((a + c)/3, b/3).

Now, the line through G parallel to BC. Since BC is horizontal, the line through G parallel to BC is also horizontal. Therefore, it has equation y = b/3. This line intersects AB and AC at points M and N. Let me find coordinates of M and N.

First, find equations of AB and AC.

Equation of AB: connects A(a, b) and B(0, 0). The parametric equations can be written as x = a*t, y = b*t, where t ranges from 0 to 1.

Similarly, equation of AC: connects A(a, b) and C(c, 0). Parametric equations can be written as x = a + (c - a)*s, y = b - b*s, where s ranges from 0 to 1.

But maybe it's easier to find the intersection points with the line y = b/3.

For AB: Let me find where y = b/3. On AB, the line from (0, 0) to (a, b). The slope of AB is (b - 0)/(a - 0) = b/a. So equation is y = (b/a)x. Setting y = b/3, we have b/3 = (b/a)x => x = a/3. Therefore, point M is at (a/3, b/3).

Similarly, for AC: The line from A(a, b) to C(c, 0). The slope is (0 - b)/(c - a) = -b/(c - a). Equation of AC: y - b = [-b/(c - a)](x - a). Let me solve for y when y = b/3.

So, b/3 - b = [-b/(c - a)](x - a)

=> -2b/3 = [-b/(c - a)](x - a)

Multiply both sides by (c - a)/(-b):

(-2b/3) * (c - a)/(-b) = x - a

Simplify:

(2/3)(c - a) = x - a

Therefore, x = a + (2/3)(c - a) = (a/3) + (2c)/3 = (a + 2c)/3

Therefore, point N is at ((a + 2c)/3, b/3)

So coordinates of M: (a/3, b/3), coordinates of N: ((a + 2c)/3, b/3). Therefore, MN is the line segment between these two points, which is horizontal at y = b/3.

Now, the problem mentions triangle ENG. Wait, but where is point E? Wait, the medians are AD, BE, and CF. So BE is a median from B to AC. So point E is the midpoint of AC. Wait, yes. Since BE is a median, E is the midpoint of AC.

Similarly, D is the midpoint of BC, and F is the midpoint of AB.

So coordinates of E: midpoint of AC. Coordinates of A(a, b) and C(c, 0), so midpoint E is ((a + c)/2, (b + 0)/2) = ((a + c)/2, b/2).

Wait, so point E is ((a + c)/2, b/2). Then point G is ((a + c)/3, b/3). So we need to find the area of triangle ENG, where N is ((a + 2c)/3, b/3), E is ((a + c)/2, b/2), and G is ((a + c)/3, b/3).

Wait, so triangle ENG has coordinates:

E: ((a + c)/2, b/2)

N: ((a + 2c)/3, b/3)

G: ((a + c)/3, b/3)

So to find the area of triangle ENG, we can use coordinates. Let me recall the formula for area given three coordinates.

The area can be calculated using determinant formula:

Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Let me assign:

E: (x1, y1) = ((a + c)/2, b/2)

N: (x2, y2) = ((a + 2c)/3, b/3)

G: (x3, y3) = ((a + c)/3, b/3)

So plug into the formula:

Area = (1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

Compute each term:

First term: x1(y2 - y3) = [(a + c)/2] * [ (b/3 - b/3) ] = [(a + c)/2] * 0 = 0

Second term: x2(y3 - y1) = [(a + 2c)/3] * [ (b/3 - b/2) ] = [(a + 2c)/3] * [ (-b/6) ] = -b(a + 2c)/18

Third term: x3(y1 - y2) = [(a + c)/3] * [ (b/2 - b/3) ] = [(a + c)/3] * [ b/6 ] = b(a + c)/18

Therefore, adding them up:

0 + (-b(a + 2c)/18) + (b(a + c)/18) = [ -b(a + 2c) + b(a + c) ] / 18 = [ -ab - 2bc + ab + bc ] / 18 = (-bc)/18

Taking absolute value and multiplying by 1/2:

Area = (1/2) * | -bc / 18 | = (1/2)(bc/18) = bc/36

So the area of triangle ENG is bc/36. Now, we need to relate this to the area of triangle ABC, which is given as 144.

What is the area of triangle ABC? Using coordinates, with A(a, b), B(0, 0), C(c, 0). The area can be calculated as (base * height)/2. The base BC is from (0, 0) to (c, 0), so length c. The height is the y-coordinate of A, which is b. Wait, but if A is at (a, b), then the base BC is along x-axis from (0, 0) to (c, 0), but point A is at (a, b). Wait, but the area of triangle ABC is |(Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By))/2|.

Plugging in coordinates:

Area = (1/2)| a*(0 - 0) + 0*(0 - b) + c*(b - 0) | = (1/2)| 0 + 0 + bc | = (1/2)| bc | = bc/2

Given that the area is 144, so bc/2 = 144 => bc = 288

Therefore, area of triangle ENG is bc/36 = 288 / 36 = 8

Wait, so the area is 8? Let me check my steps again to make sure.

First, coordinates assigned: A(a, b), B(0, 0), C(c, 0). Then centroid G is ((a + c)/3, b/3). Then line through G parallel to BC (which is horizontal) is y = b/3. Found points M and N on AB and AC. Calculated coordinates of M as (a/3, b/3) and N as ((a + 2c)/3, b/3). Then point E is midpoint of AC: ((a + c)/2, b/2). Then coordinates of E, N, G. Then using determinant formula, found area of ENG as bc/36. Then area of ABC is bc/2 = 144 => bc = 288. Therefore, bc/36 = 8. So area of ENG is 8. That seems straightforward. But let me check if there's a mistake in coordinates.

Wait, but in my coordinate system, point A is (a, b), but if I set coordinates with B at (0, 0) and C at (c, 0), then AB is from (0, 0) to (a, b). However, if I want to make sure that the coordinates are general, maybe I need to check if there are any constraints on a? For example, if I take AB from (0, 0) to (a, b), then AC is from (a, b) to (c, 0). But maybe there's a better coordinate system where calculations are simpler?

Alternatively, maybe I can assign coordinates such that BC is of length 2 units or something? Wait, but since the area is given as 144, maybe it's better to assign coordinates such that calculations are easier. Let me try another approach.

Alternatively, let me take specific coordinates for triangle ABC. Let me assume coordinates where BC is horizontal, and centroid is at (0, 0). Wait, but maybe that complicates things. Alternatively, take coordinates where A is at (0, 0), B at (2, 0), C at (0, 2). Wait, but then area would be 2, but given area is 144. Maybe scaling later?

Wait, but maybe using vectors or ratios?

Alternatively, since the area of ABC is 144, and we found that area of ENG is bc/36, and bc/2 = 144, so bc = 288, so bc/36 = 8. Therefore, answer is 8? But let me check with specific coordinates.

Let me take specific values. Let me choose coordinates where A is (0, 6), B is (0, 0), C is (6, 0). Then area of ABC is (base * height)/2 = (6 * 6)/2 = 18. Wait, but I need area 144. Let me scale up. Let me take A(0, 12), B(0, 0), C(12, 0). Then area is (12 * 12)/2 = 72. Still not 144. Let me take A(0, 24), B(0, 0), C(12, 0). Then area is (12 * 24)/2 = 144. Perfect. So let me use this coordinate system.

So A(0, 24), B(0, 0), C(12, 0). Then centroid G is average of coordinates: ((0 + 0 + 12)/3, (24 + 0 + 0)/3) = (4, 8).

Now, line through G parallel to BC. Since BC is horizontal (from (0,0) to (12, 0)), so line through G(4, 8) parallel to BC is horizontal line y = 8. This line intersects AB and AC at M and N.

First, find equation of AB: from (0, 24) to (0, 0). Wait, AB is vertical line x = 0? Wait, but if A is (0, 24) and B is (0, 0), then AB is vertical. Then line y = 8 intersects AB at (0, 8). Wait, but the problem says the line through G parallel to BC intersects AB and AC at M and N. But if AB is vertical, then the line y = 8 intersects AB at (0, 8), which is point M. Then where does it intersect AC?

Equation of AC: from A(0, 24) to C(12, 0). The slope is (0 - 24)/(12 - 0) = -2. Equation is y - 24 = -2x => y = -2x + 24. Find intersection with y = 8:

8 = -2x + 24 => -2x = -16 => x = 8. Therefore, point N is (8, 8). Therefore, points M(0, 8) and N(8, 8).

Now, point E is the midpoint of AC. Coordinates of A(0, 24) and C(12, 0). Midpoint E is ((0 + 12)/2, (24 + 0)/2) = (6, 12).

So triangle ENG has points E(6, 12), N(8, 8), G(4, 8). Let me compute its area.

Using coordinates:

E(6, 12), N(8, 8), G(4, 8).

Using the determinant formula:

Area = (1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

Plugging in:

x1 = 6, y1 = 12

x2 = 8, y2 = 8

x3 = 4, y3 = 8

Compute:

6*(8 - 8) + 8*(8 - 12) + 4*(12 - 8)

= 6*0 + 8*(-4) + 4*(4)

= 0 - 32 + 16 = -16

Take absolute value and multiply by 1/2: (1/2)*| -16 | = 8.

So area is 8. Which matches the previous result. Since in this coordinate system, area of ABC is 144, and area of ENG is 8. Therefore, the answer is 8.

But let me check with another coordinate system to be sure. Let me take another triangle. Let me take A(0, 0), B(6, 0), C(0, 8). Then area of ABC is (6*8)/2 = 24. Let me compute area of ENG in this case, then see if the ratio holds.

First, centroid G is ((0 + 6 + 0)/3, (0 + 0 + 8)/3) = (2, 8/3).

Line through G parallel to BC. BC is from (6, 0) to (0, 8). Wait, BC is not horizontal here. Wait, in this coordinate system, BC is from B(6, 0) to C(0, 8). So slope of BC is (8 - 0)/(0 - 6) = -4/3. Therefore, line through G parallel to BC would have slope -4/3.

But the problem says the line through G parallel to BC intersects AB and AC at M and N. Wait, but in this coordinate system, AB is from A(0, 0) to B(6, 0), which is horizontal. AC is from A(0, 0) to C(0, 8), vertical. So line through G(2, 8/3) with slope -4/3. Let me find where this line intersects AB and AC.

First, equation of the line: y - 8/3 = (-4/3)(x - 2)

Simplify: y = (-4/3)x + 8/3 + 8/3 = (-4/3)x + 16/3

Intersection with AB: AB is y = 0. So set y = 0:

0 = (-4/3)x + 16/3 => (-4/3)x = -16/3 => x = 4. Therefore, point M is (4, 0).

Intersection with AC: AC is x = 0. So plug x = 0 into line equation:

y = (-4/3)(0) + 16/3 = 16/3. Therefore, point N is (0, 16/3).

Now, point E is the midpoint of AC. AC is from (0, 0) to (0, 8), so midpoint E is (0, 4).

So triangle ENG has points E(0, 4), N(0, 16/3), G(2, 8/3). Let me compute its area.

Using coordinates:

E(0, 4), N(0, 16/3), G(2, 8/3)

Using determinant formula:

Area = (1/2)| x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |

Plugging in:

x1 = 0, y1 = 4

x2 = 0, y2 = 16/3

x3 = 2, y3 = 8/3

Compute:

0*(16/3 - 8/3) + 0*(8/3 - 4) + 2*(4 - 16/3)

= 0 + 0 + 2*(12/3 - 16/3) = 2*(-4/3) = -8/3

Absolute value times 1/2: (1/2)*(8/3) = 4/3

Now, area of ABC is 24. What is bc/36 in this case? Wait, in my previous coordinate system, I had bc/36. Wait, but in this coordinate system, how does that relate?

Wait, in my first general coordinate system, I had A(a, b), B(0, 0), C(c, 0). Then area of ABC was bc/2. In this case, A(0, 0), B(6, 0), C(0, 8). So here, if I consider similar coordinates, but A is at (0, 0), B at (c, 0), C at (0, b). Then area is (c*b)/2. In this case, c = 6, b = 8, so area is 24, which matches. Then bc/36 would be (6*8)/36 = 48/36 = 4/3, which matches the area of triangle ENG here. Therefore, in this case, area of ENG is 4/3, and area of ABC is 24, so ratio is (4/3)/24 = 1/18. Therefore, in general, area of ENG is (area ABC)/18. Since in the first case, area ABC was 144, then area ENG is 144 / 18 = 8. Which matches. Therefore, the answer is 8.

Therefore, regardless of the coordinate system, the area of triangle ENG is 1/18 of the area of ABC. Therefore, since area ABC is 144, area ENG is 144 / 18 = 8.

But let me think again why the ratio is 1/18. In the first coordinate system, bc/36 divided by bc/2 is (bc/36)/(bc/2) = 2/36 = 1/18. So yes, the ratio is 1/18. Therefore, regardless of the triangle, the area of ENG is 1/18 of ABC. Therefore, answer is 144 / 18 = 8.

Alternatively, maybe there's a way to see this without coordinates? Let me think.

Since MN is parallel to BC and passes through centroid G. Since centroid divides the median in ratio 2:1, then the distance from G to BC is 1/3 the height of the triangle. Wait, if we consider the line MN parallel to BC through G, then the distance between MN and BC is 1/3 the height? Wait, no. Wait, the centroid is located at 1/3 of the height from the base. Wait, if we consider the height from A to BC, then centroid is located 1/3 of the height from BC. Therefore, the line MN is located at 1/3 the height from BC, so the distance between MN and BC is 1/3 the height. Therefore, the ratio of areas between triangle AMN and ABC would be (2/3)^2 = 4/9? Wait, no. Wait, if a line is drawn parallel to the base at a fraction k of the height, then the ratio of areas is k^2. But here, MN is located at 1/3 the height from BC, so the distance from A to MN is 2/3 the height. Therefore, the ratio of areas of AMN to ABC is (2/3)^2 = 4/9. Therefore, area of AMN is 4/9 * 144 = 64. Then area of the trapezoid MNBC is 144 - 64 = 80. But how does this relate to triangle ENG?

Alternatively, maybe considering coordinates again. Since MN is parallel to BC and passes through G, which divides the medians in ratio 2:1. Maybe using mass point or area ratios.

Alternatively, since E is the midpoint of AC, and N is a point on AC. Wait, in the coordinates earlier, N was located at ((a + 2c)/3, b/3). So in terms of AC, which goes from A(a, b) to C(c, 0). The point N divides AC in what ratio? Let me compute.

Parametrize AC from A to C. Let me use parameter t. Coordinates of N are ((a + 2c)/3, b/3). So starting at A(a, b), moving towards C(c, 0). The x-coordinate goes from a to c, so the difference is c - a. The x-coordinate of N is a + (c - a)*(2/3). Therefore, it's 2/3 of the way from A to C? Wait, no. Wait, if t is the fraction from A to C, then x = a + t(c - a). So solving for t:

a + t(c - a) = (a + 2c)/3

=> t(c - a) = (a + 2c)/3 - a = (a + 2c - 3a)/3 = (-2a + 2c)/3 = 2(c - a)/3

Therefore, t = [2(c - a)/3] / (c - a) ) = 2/3. Therefore, point N divides AC in ratio AN:NC = 2:1. Similarly, point M divides AB in ratio AM:MB = 2:1? Wait, in coordinates earlier, point M was at (a/3, b/3). If AB is from A(a, b) to B(0, 0). So parametrize AB with parameter t, x = a - a*t, y = b - b*t. Wait, but earlier we found that when y = b/3, t = 1/3. Therefore, point M divides AB in ratio AM:MB = 1:2. Wait, because t = 1/3 from A to B? Wait, if t is 1/3, then AM is 1/3 of AB, so MB is 2/3 of AB. Therefore, AM:MB = 1:2. Similarly, AN:NC = 2:1.

Therefore, MN is a line connecting points dividing AB and AC in ratios 1:2 and 2:1 respectively. Therefore, MN is a line segment inside the triangle.

Now, triangle ENG. Let me think about the positions of E, N, G. E is midpoint of AC, N divides AC in ratio 2:1, G is centroid.

Alternatively, since we know coordinates, maybe using vectors?

Let me consider vectors. Let me take A as origin. Let me denote vectors AB as vector b, and AC as vector c. Then coordinates:

- Point A: 0

- Point B: b

- Point C: c

- Centroid G: (A + B + C)/3 = (0 + b + c)/3 = (b + c)/3

- Point E is midpoint of AC: (A + C)/2 = (0 + c)/2 = c/2

- Point N is on AC such that AN:NC = 2:1. Therefore, vector AN = (2/3)AC = (2/3)c. Therefore, position vector of N is A + AN = 0 + (2/3)c = (2/3)c

- Point M is on AB such that AM:MB = 1:2. Therefore, vector AM = (1/3)AB = (1/3)b. Therefore, position vector of M is (1/3)b

Now, triangle ENG has points E, N, G. Let me find vectors for these points:

- E: c/2

- N: (2/3)c

- G: (b + c)/3

So vectors from E to N and E to G?

Wait, to compute area of triangle ENG, we can use cross product. In vector terms, area is (1/2)| (N - E) × (G - E) |.

Compute N - E = (2/3)c - (1/2)c = (4/6 - 3/6)c = (1/6)c

Compute G - E = (b + c)/3 - c/2 = (2(b + c) - 3c)/6 = (2b + 2c - 3c)/6 = (2b - c)/6

Then cross product (N - E) × (G - E) = (1/6)c × (2b - c)/6 = (1/36)(c × 2b - c × c) = (1/36)(2c × b - 0) = (2/36)(c × b) = (1/18)(c × b)

The magnitude of this cross product is (1/18)|c × b|. The area of triangle ENG is half of that, so (1/36)|c × b|.

But the area of triangle ABC is (1/2)|b × c|. Therefore, area of ENG is (1/36)|c × b| = (1/18)*(1/2)|b × c| = (1/18)*Area ABC. Therefore, area of ENG is 1/18 of area ABC. Since area ABC is 144, area ENG is 144 / 18 = 8.

Therefore, regardless of coordinate system, the area is 8.

This confirms the previous results. Therefore, the answer is 8.

But just to make sure, let me think if there's another way someone might approach this problem and maybe get confused. For example, someone might think that since MN is parallel to BC and passes through centroid, then maybe triangle AMN has area 4/9 of ABC, but then how does ENG relate? But in our case, ENG is a smaller triangle inside AMN or something else. But through coordinates and vectors, we saw that it's 1/18.

Alternatively, using area ratios. Since E is midpoint of AC, and N divides AC in ratio 2:1, so EN is a segment from midpoint to a point 2/3 along AC. Then, maybe considering coordinates or mass point. But since we already have multiple methods leading to 8, I think it's safe.

Therefore, the area of triangle ENG is 8.

**Final Answer**
\boxed{8}

To solve the problem, we are given a triangle $ ABC $ with medians $ AD $, $ BE $, and $ CF $ intersecting at the centroid $ G $. A line through $ G $, parallel to $ BC $, intersects $ AB $ and $ AC $ at points $ M $ and $ N $, respectively. The area of triangle $ ABC $ is 144, and we are to find the area of triangle $ ENG $, where $ E $ is the midpoint of $ AC $.

---

### Step 1: Understanding the Geometry

- The **centroid** $ G $ divides each median in the ratio $ 2:1 $, with the longer segment closer to the vertex.
- The line through $ G $ parallel to $ BC $ intersects $ AB $ at $ M $ and $ AC $ at $ N $.
- Since the line is parallel to $ BC $, triangle $ AMN $ is similar to triangle $ ABC $, and the line passes through the centroid, which lies $ \frac{1}{3} $ of the way from the base $ BC $ to the vertex $ A $.
- Therefore, the height of triangle $ AMN $ is $ \frac{2}{3} $ of the height of triangle $ ABC $, and the area of triangle $ AMN $ is $ \left(\frac{2}{3}\right)^2 = \frac{4}{9} $ of the area of triangle $ ABC $.

---

### Step 2: Coordinates and Area Calculation

To compute the area of triangle $ ENG $, we can use coordinate geometry. Let’s place triangle $ ABC $ in the coordinate plane:

- Let $ A = (0, b) $, $ B = (0, 0) $, and $ C = (c, 0) $.
- Then the area of triangle $ ABC $ is:
  $$
  \text{Area}_{ABC} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot c \cdot b = \frac{bc}{2}
  $$
  Given that this area is 144, we have:
  $$
  \frac{bc}{2} = 144 \Rightarrow bc = 288
  $$

Now, we find the coordinates of the relevant points:

- **Centroid $ G $**:
  $$
  G = \left( \frac{0 + 0 + c}{3}, \frac{b + 0 + 0}{3} \right) = \left( \frac{c}{3}, \frac{b}{3} \right)
  $$

- **Point $ E $**, the midpoint of $ AC $:
  $$
  E = \left( \frac{0 + c}{2}, \frac{b + 0}{2} \right) = \left( \frac{c}{2}, \frac{b}{2} \right)
  $$

- **Point $ N $**, the intersection of the line through $ G $ parallel to $ BC $ with $ AC $:
  Since the line is horizontal at $ y = \frac{b}{3} $, and $ AC $ has slope $ -\frac{b}{c} $, we find the x-coordinate of $ N $ by solving:
  $$
  y = -\frac{b}{c}x + b = \frac{b}{3} \Rightarrow x = \frac{2c}{3}
  $$
  So, $ N = \left( \frac{2c}{3}, \frac{b}{3} \right) $

Now, we have the coordinates of triangle $ ENG $:

- $ E = \left( \frac{c}{2}, \frac{b}{2} \right) $
- $ N = \left( \frac{2c}{3}, \frac{b}{3} \right) $
- $ G = \left( \frac{c}{3}, \frac{b}{3} \right) $

---

### Step 3: Area of Triangle $ ENG $

Using the determinant formula for the area of a triangle with vertices $ (x_1, y_1) $, $ (x_2, y_2) $, $ (x_3, y_3) $:

$$
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
$$

Substitute the coordinates:

- $ x_1 = \frac{c}{2}, y_1 = \frac{b}{2} $
- $ x_2 = \frac{2c}{3}, y_2 = \frac{b}{3} $
- $ x_3 = \frac{c}{3}, y_3 = \frac{b}{3} $

Compute each term:

- $ x_1(y_2 - y_3) = \frac{c}{2} \cdot \left( \frac{b}{3} - \frac{b}{3} \right) = 0 $
- $ x_2(y_3 - y_1) = \frac{2c}{3} \cdot \left( \frac{b}{3} - \frac{b}{2} \right) = \frac{2c}{3} \cdot \left( -\frac{b}{6} \right) = -\frac{bc}{9} $
- $ x_3(y_1 - y_2) = \frac{c}{3} \cdot \left( \frac{b}{2} - \frac{b}{3} \right) = \frac{c}{3} \cdot \frac{b}{6} = \frac{bc}{18} $

Add them:

$$
\text{Area} = \frac{1}{2} \left| 0 - \frac{bc}{9} + \frac{bc}{18} \right| = \frac{1}{2} \left| -\frac{bc}{18} \right| = \frac{1}{2} \cdot \frac{bc}{18} = \frac{bc}{36}
$$

We already know $ bc = 288 $, so:

$$
\text{Area}_{ENG} = \frac{288}{36} = 8
$$

---

### Final Answer

$$
\boxed{8}
$$

Solution:
Since $E$ is the midpoint of $AC$, the area of triangle $BCE$ is half the area of triangle $ABC$, or $144/2 = 72$.

[asy]
import geometry;

unitsize(1 cm);

pair A, B, C, D, E, F, G, M, N;

A = (1,3);
B = (0,0);
C = (4,0);
D = (B + C)/2;
E = (C + A)/2;
F = (A + B)/2;
G = (A + B + C)/3;
M = extension(G, G + B - C, A, B);
N = extension(G, G + B - C, A, C);

draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);
draw(M--N);

label("$A$", A, dir(90));
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$G$", G, SSW);
label("$M$", M, NW);
label("$N$", N, NE);
[/asy]

Since $GN$ is parallel to $BC$, triangles $ENG$ and $ECB$ are similar.  Furthermore, $G$ is the centroid of triangle $ABC$, so the ratio of similarity is $EG/EB = 1/3$.  Therefore, the area of triangle $ENG$ is $72 \cdot (1/3)^2 = \boxed{8}$.