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. The problem also gives me three equations involving the distances between the points (x1,y1), (x2,y2), and (x3,y3).

First, let me recall that the determinant of a 3x3 matrix with columns (x1, y1, 1), (x2, y2, 1), (x3, y3, 1) is related to the area of the triangle formed by the three points. Wait, but the determinant formula for area... actually, the area of the triangle formed by three points (x1,y1), (x2,y2), (x3,y3) is given by half the absolute value of the determinant of the matrix:

$$
\frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|
$$

But in this problem, they are asking for the square of the determinant, not the absolute value. However, since determinant can be positive or negative, but squared, it would be the same as the square of the absolute value. So, if I can find the area, then multiply by 2 and then square it, but maybe there's a direct way.

But maybe I need to relate the given distances to the determinant. Let me think. The three equations given are the squares of the distances between each pair of points. Let me denote the points as A(x1, y1), B(x2, y2), and C(x3, y3). Then:

AB² = 9, AC² = 16, BC² = 25. So, AB = 3, AC = 4, BC = 5. Wait, 3-4-5 triangle? That's a right triangle! Since 3² + 4² = 5². So triangle ABC is a right triangle with right angle at A? Let me check: AB² + AC² = 9 + 16 = 25 = BC², which is the Pythagorean theorem. So yes, triangle ABC is right-angled at A. That might be useful.

Now, for a right-angled triangle, the area is (AB * AC)/2 = (3*4)/2 = 6. Therefore, the area of triangle ABC is 6. But earlier I mentioned that the area is half the absolute value of the determinant. Hence:

$$
\frac{1}{2} \left| \text{det} \right| = 6 \implies \left| \text{det} \right| = 12 \implies \text{det}^2 = 144
$$

Wait, so is the answer just 144? But let me check if this is correct. Because maybe I need to be careful with the determinant.

Wait, the determinant in question is:

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

Which is equal to the determinant of the matrix formed by the coordinates of the three points as columns, with an extra column of ones. However, the formula for the area is indeed half the absolute value of this determinant. But let me confirm by expanding the determinant.

Expanding the determinant:

Let me denote the determinant as D. Then,

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

Simplify:

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

Which can be written as:

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

Hmm, maybe there's another way to write this. Alternatively, note that the determinant is twice the area of the triangle. Wait, no, I thought it was half. Let me recall.

The area can be calculated using the shoelace formula, which is similar to the determinant. If we have points A(x1, y1), B(x2, y2), C(x3, y3), then the area is:

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

Which is similar to the determinant. Let me check:

Expanding the determinant:

Using the rule of Sarrus or cofactor expansion. Let me compute it properly.

The determinant D:

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

Expanding along the third column (since it has all ones, which might be easier):

D = 1*(x1y2 - x2y1) - 1*(x1y3 - x3y1) + 1*(x2y3 - x3y2)

Simplify:

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

Grouping terms:

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

Rearranged:

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

Which is the same as the shoelace formula expression. But the shoelace formula is:

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

So indeed, D = 2 * Area. Therefore, the absolute value of D is 2 * Area. Therefore, Area = |D| / 2. Therefore, if the area is 6, then |D| = 12, so D² = 144. Therefore, the answer is 144?

But wait, but let me make sure that triangle ABC is indeed right-angled. Because if AB² + AC² = BC², then the right angle is at A. So angle at A is right angle. Therefore, area is 1/2 AB * AC = 1/2 * 3 * 4 = 6. Therefore, yes, so |D| is 12, so D squared is 144.

But wait, the problem didn't specify the order of the points. Wait, but the three equations are:

(x1 - x2)^2 + (y1 - y2)^2 = 9,

(x1 - x3)^2 + (y1 - y3)^2 = 16,

(x2 - x3)^2 + (y2 - y3)^2 = 25.

So AB = 3, AC = 4, BC = 5, so triangle with sides 3,4,5. So right-angled at A. So regardless of the order? Wait, if AB = 3, AC = 4, BC = 5, then AB² + AC² = BC², so angle at A is right. Therefore, the area is 6. Therefore, determinant squared is 144.

But let me check if there is another way to approach this, just to confirm. Maybe using coordinates?

Alternatively, suppose we assign coordinates to the points. Let me assume point A is at (0,0), since we can translate the coordinate system. Since it's a right triangle at A, then let me set A(0,0), B(3,0), C(0,4). Then the distances: AB is 3 units, AC is 4 units, BC would be 5 units. Then compute determinant.

So let me compute determinant for points A(0,0), B(3,0), C(0,4):

Matrix is:

[0, 0, 1]

[3, 0, 1]

[0, 4, 1]

Compute determinant:

0*(0*1 - 4*1) - 0*(3*1 - 0*1) + 1*(3*4 - 0*0) = 0 - 0 + 1*(12 - 0) = 12. Therefore determinant is 12, so squared is 144. So same answer. Therefore, regardless of coordinate system, the determinant squared is 144. Therefore, answer is 144.

But wait, but the problem didn't specify coordinates, but since the distances are given, and the determinant is up to translation, rotation, etc. But determinant is dependent on coordinates. However, since the area is fixed (as we found 6), then determinant squared is fixed. Therefore, answer is 144.

But maybe there's a case where determinant could be negative? But since we square it, it's still 144. Therefore, the answer is 144.

Alternatively, maybe I can think in terms of vectors. Let me try another approach.

Let me denote vectors AB and AC. Since AB = 3, AC = 4, and they are perpendicular. Then the area is 1/2 |AB × AC| = 6. The cross product in 2D can be represented as AB_x * AC_y - AB_y * AC_x. But the determinant D is related to the area. Wait, if we have vectors AB and AC, then the determinant of the matrix with columns AB and AC is the area of the parallelogram, which is twice the area of the triangle. But in our case, the determinant D is different.

Alternatively, maybe express the determinant in terms of vectors. Let me think. The determinant D is equal to the area of the parallelogram formed by vectors from A to B and from A to C multiplied by something? Wait, earlier we saw that determinant D is twice the area of triangle ABC. So regardless, since area of triangle is 6, determinant is 12.

Alternatively, maybe use coordinates in general. Suppose we have three points with mutual distances 3,4,5. Then, regardless of their positions, the determinant squared is fixed? Because the area is fixed. So even if you rotate or translate the triangle, the determinant squared would remain the same? Wait, determinant depends on coordinates, but if you translate the points, determinant changes? Wait, no. Wait, the determinant is dependent on the coordinates. Let me check with translation.

Suppose I translate all points by some vector (h, k). Then the determinant would change, right? Because each column is translated. But the area remains the same. Wait, but determinant is twice the area, so determinant would be the same? Wait, no. Let me check with an example.

Suppose original points A(0,0), B(3,0), C(0,4). Then determinant is 12. Now, translate all points by (h,k): A'(h,k), B'(h+3, k), C'(h, k+4). Compute determinant:

|h  k   1|

|h+3 k  1|

|h  k+4 1|

Calculating determinant:

h*(k*1 - (k+4)*1) - k*( (h+3)*1 - h*1 ) + 1*( (h+3)(k+4) - h*k )

Simplify:

h*(k - k -4) - k*(h + 3 - h) + [ (h+3)(k+4) - h k ]

= h*(-4) - k*(3) + [hk + 4h + 3k + 12 - hk]

= -4h - 3k + [4h + 3k + 12]

= (-4h -3k) + 4h + 3k + 12 = 12.

So determinant is still 12, regardless of translation. Therefore, determinant is translation invariant? Wait, but the determinant is 12 in this case. So even when translated, determinant remains same. So that's interesting. So determinant is invariant under translation. Therefore, no matter how we translate the triangle, determinant remains same. Therefore, even if we rotate the triangle, would determinant change? Let me check with rotation.

Suppose we rotate the triangle. Let me take point A(0,0), B(3,0), C(0,4). Rotate by 90 degrees around origin. Then new points: A'(0,0), B'(0,3), C'(-4,0). Compute determinant:

[0  0  1]

[0  3  1]

[-4 0 1]

Determinant:

0*(3*1 - 0*1) - 0*(0*1 - (-4)*1) + 1*(0*0 - (-4)*3) = 0 - 0 + 1*(0 + 12) = 12. Same determinant. So determinant is invariant under rotation. Therefore, determinant is same regardless of rotation and translation. Therefore, regardless of coordinate system, determinant is 12, so squared is 144. Therefore, answer is 144.

Therefore, regardless of the positions of the points, as long as they form a triangle with sides 3,4,5, determinant is 12, square is 144. Therefore, the answer is 144.

But let me check if there's a case where the determinant could be different. Suppose the triangle is not right-angled. Wait, but in our case, the distances are 3,4,5, which is fixed. So regardless of how the triangle is placed, the area is fixed. Therefore, determinant is fixed. Therefore, answer is 144.

But maybe to make sure, let me think of another configuration. Suppose points are not placed in the coordinate system as I did. Let me assign coordinates differently. Suppose I have points A(0,0), B(0,3), C(4,0). Wait, this is same as before, but swapping coordinates. Then the determinant would be:

[0  0  1]

[0  3  1]

[4  0  1]

Determinant:

0*(3*1 - 0*1) - 0*(0*1 - 4*1) + 1*(0*0 - 4*3) = 0 - 0 + 1*(0 - 12) = -12. Absolute value 12, square is 144. Same result.

Alternatively, suppose I have different coordinates where the triangle is placed in a different way. Let me say point A(1,1), B(1,4), C(5,1). Let me check distances:

AB: sqrt( (1-1)^2 + (4-1)^2 ) = 3, good.

AC: sqrt( (5-1)^2 + (1-1)^2 ) = 4, good.

BC: sqrt( (5-1)^2 + (1-4)^2 ) = sqrt(16 + 9) = 5, good.

Now compute determinant:

[1  1  1]

[1  4  1]

[5  1  1]

Determinant calculation:

1*(4*1 - 1*1) - 1*(1*1 - 5*1) + 1*(1*1 - 5*4)

= 1*(4 - 1) - 1*(1 - 5) + 1*(1 - 20)

= 1*3 - 1*(-4) + 1*(-19)

= 3 + 4 -19 = -12. Absolute value 12, square 144. Same result.

Therefore, regardless of coordinates, determinant squared is 144. Therefore, answer is 144.

But let me think again if there is a possibility that the triangle is not right-angled. Wait, given AB² + AC² = BC², so it must be right-angled. Therefore, area is fixed. Therefore determinant squared is fixed. Hence, the answer is 144. Therefore, I can confidently say that the answer is 144.

Wait, but let me think again if there's another approach, maybe using coordinates algebraically. Let me try to compute the determinant squared using the given distances. Maybe using the formula for determinant squared in terms of the squares of distances?

Alternatively, I can recall that the square of the determinant is related to the area squared, which is related to the sides. But since we have the sides, maybe using Heron's formula? Let me try that.

For triangle with sides a=3, b=4, c=5. Area is 6, as we know. So Heron's formula gives:

s = (a + b + c)/2 = (3+4+5)/2 = 6.

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. Correct. Therefore, area is 6. Therefore, determinant squared is (2 * area)^2 = (12)^2 = 144. Same result.

Alternatively, perhaps we can relate the determinant squared with the given squared distances. But seems more complicated. However, just to explore, maybe using coordinates.

Let me denote the determinant as D. Then D^2 = (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))^2.

But maybe expanding this expression? But perhaps complicated. However, given that we already have a solid answer through area, perhaps not necessary.

Alternatively, think of the determinant squared as the square of the area times 4. Since D = 2 * area, so D² = 4 * area². Since area is 6, then 4*36 = 144.

Therefore, answer is 144.

Therefore, I think the answer is 144, so boxed answer is \boxed{144}.

**Final Answer**
\boxed{144}

To solve the problem, we are given three equations representing 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 distances correspond to the lengths of the sides of a triangle with points $ A(x_1, y_1) $, $ B(x_2, y_2) $, and $ C(x_3, y_3) $, denoted as:

- $ AB = 3 $,
- $ AC = 4 $,
- $ BC = 5 $.

We observe that:

$$
AB^2 + AC^2 = 9 + 16 = 25 = BC^2
$$

This confirms that triangle $ ABC $ is a **right triangle** with a right angle at $ A $.

---

### Area of the Triangle

For a right triangle with legs $ AB = 3 $ and $ AC = 4 $, the area is given by:

$$
\text{Area} = \frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} \cdot 3 \cdot 4 = 6
$$

However, the determinant we are asked to consider is the determinant of the following matrix:

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

This determinant, when expanded, represents **twice the signed area** of the triangle $ ABC $ (up to a sign). So,

$$
\left| \begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix} \right| = 2 \times \text{Area} = 2 \times 6 = 12
$$

Therefore, the **square of the absolute value of the determinant** is:

$$
\left(\left| \begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix} \right| \right)^2 = 12^2 = \boxed{144}
$$

---

### 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. The problem also gives me three equations involving the distances between the points (x1,y1), (x2,y2), and (x3,y3).

First, let me recall that the determinant of a 3x3 matrix with columns (x1, y1, 1), (x2, y2, 1), (x3, y3, 1) is related to the area of the triangle formed by the three points. Wait, but the determinant formula for area... actually, the area of the triangle formed by three points (x1,y1), (x2,y2), (x3,y3) is given by half the absolute value of the determinant of the matrix:

$$
\frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|
$$

But in this problem, they are asking for the square of the determinant, not the absolute value. However, since determinant can be positive or negative, but squared, it would be the same as the square of the absolute value. So, if I can find the area, then multiply by 2 and then square it, but maybe there's a direct way.

But maybe I need to relate the given distances to the determinant. Let me think. The three equations given are the squares of the distances between each pair of points. Let me denote the points as A(x1, y1), B(x2, y2), and C(x3, y3). Then:

AB² = 9, AC² = 16, BC² = 25. So, AB = 3, AC = 4, BC = 5. Wait, 3-4-5 triangle? That's a right triangle! Since 3² + 4² = 5². So triangle ABC is a right triangle with right angle at A? Let me check: AB² + AC² = 9 + 16 = 25 = BC², which is the Pythagorean theorem. So yes, triangle ABC is right-angled at A. That might be useful.

Now, for a right-angled triangle, the area is (AB * AC)/2 = (3*4)/2 = 6. Therefore, the area of triangle ABC is 6. But earlier I mentioned that the area is half the absolute value of the determinant. Hence:

$$
\frac{1}{2} \left| \text{det} \right| = 6 \implies \left| \text{det} \right| = 12 \implies \text{det}^2 = 144
$$

Wait, so is the answer just 144? But let me check if this is correct. Because maybe I need to be careful with the determinant.

Wait, the determinant in question is:

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

Which is equal to the determinant of the matrix formed by the coordinates of the three points as columns, with an extra column of ones. However, the formula for the area is indeed half the absolute value of this determinant. But let me confirm by expanding the determinant.

Expanding the determinant:

Let me denote the determinant as D. Then,

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

Simplify:

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

Which can be written as:

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

Hmm, maybe there's another way to write this. Alternatively, note that the determinant is twice the area of the triangle. Wait, no, I thought it was half. Let me recall.

The area can be calculated using the shoelace formula, which is similar to the determinant. If we have points A(x1, y1), B(x2, y2), C(x3, y3), then the area is:

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

Which is similar to the determinant. Let me check:

Expanding the determinant:

Using the rule of Sarrus or cofactor expansion. Let me compute it properly.

The determinant D:

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

Expanding along the third column (since it has all ones, which might be easier):

D = 1*(x1y2 - x2y1) - 1*(x1y3 - x3y1) + 1*(x2y3 - x3y2)

Simplify:

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

Grouping terms:

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

Rearranged:

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

Which is the same as the shoelace formula expression. But the shoelace formula is:

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

So indeed, D = 2 * Area. Therefore, the absolute value of D is 2 * Area. Therefore, Area = |D| / 2. Therefore, if the area is 6, then |D| = 12, so D² = 144. Therefore, the answer is 144?

But wait, but let me make sure that triangle ABC is indeed right-angled. Because if AB² + AC² = BC², then the right angle is at A. So angle at A is right angle. Therefore, area is 1/2 AB * AC = 1/2 * 3 * 4 = 6. Therefore, yes, so |D| is 12, so D squared is 144.

But wait, the problem didn't specify the order of the points. Wait, but the three equations are:

(x1 - x2)^2 + (y1 - y2)^2 = 9,

(x1 - x3)^2 + (y1 - y3)^2 = 16,

(x2 - x3)^2 + (y2 - y3)^2 = 25.

So AB = 3, AC = 4, BC = 5, so triangle with sides 3,4,5. So right-angled at A. So regardless of the order? Wait, if AB = 3, AC = 4, BC = 5, then AB² + AC² = BC², so angle at A is right. Therefore, the area is 6. Therefore, determinant squared is 144.

But let me check if there is another way to approach this, just to confirm. Maybe using coordinates?

Alternatively, suppose we assign coordinates to the points. Let me assume point A is at (0,0), since we can translate the coordinate system. Since it's a right triangle at A, then let me set A(0,0), B(3,0), C(0,4). Then the distances: AB is 3 units, AC is 4 units, BC would be 5 units. Then compute determinant.

So let me compute determinant for points A(0,0), B(3,0), C(0,4):

Matrix is:

[0, 0, 1]

[3, 0, 1]

[0, 4, 1]

Compute determinant:

0*(0*1 - 4*1) - 0*(3*1 - 0*1) + 1*(3*4 - 0*0) = 0 - 0 + 1*(12 - 0) = 12. Therefore determinant is 12, so squared is 144. So same answer. Therefore, regardless of coordinate system, the determinant squared is 144. Therefore, answer is 144.

But wait, but the problem didn't specify coordinates, but since the distances are given, and the determinant is up to translation, rotation, etc. But determinant is dependent on coordinates. However, since the area is fixed (as we found 6), then determinant squared is fixed. Therefore, answer is 144.

But maybe there's a case where determinant could be negative? But since we square it, it's still 144. Therefore, the answer is 144.

Alternatively, maybe I can think in terms of vectors. Let me try another approach.

Let me denote vectors AB and AC. Since AB = 3, AC = 4, and they are perpendicular. Then the area is 1/2 |AB × AC| = 6. The cross product in 2D can be represented as AB_x * AC_y - AB_y * AC_x. But the determinant D is related to the area. Wait, if we have vectors AB and AC, then the determinant of the matrix with columns AB and AC is the area of the parallelogram, which is twice the area of the triangle. But in our case, the determinant D is different.

Alternatively, maybe express the determinant in terms of vectors. Let me think. The determinant D is equal to the area of the parallelogram formed by vectors from A to B and from A to C multiplied by something? Wait, earlier we saw that determinant D is twice the area of triangle ABC. So regardless, since area of triangle is 6, determinant is 12.

Alternatively, maybe use coordinates in general. Suppose we have three points with mutual distances 3,4,5. Then, regardless of their positions, the determinant squared is fixed? Because the area is fixed. So even if you rotate or translate the triangle, the determinant squared would remain the same? Wait, determinant depends on coordinates, but if you translate the points, determinant changes? Wait, no. Wait, the determinant is dependent on the coordinates. Let me check with translation.

Suppose I translate all points by some vector (h, k). Then the determinant would change, right? Because each column is translated. But the area remains the same. Wait, but determinant is twice the area, so determinant would be the same? Wait, no. Let me check with an example.

Suppose original points A(0,0), B(3,0), C(0,4). Then determinant is 12. Now, translate all points by (h,k): A'(h,k), B'(h+3, k), C'(h, k+4). Compute determinant:

|h  k   1|

|h+3 k  1|

|h  k+4 1|

Calculating determinant:

h*(k*1 - (k+4)*1) - k*( (h+3)*1 - h*1 ) + 1*( (h+3)(k+4) - h*k )

Simplify:

h*(k - k -4) - k*(h + 3 - h) + [ (h+3)(k+4) - h k ]

= h*(-4) - k*(3) + [hk + 4h + 3k + 12 - hk]

= -4h - 3k + [4h + 3k + 12]

= (-4h -3k) + 4h + 3k + 12 = 12.

So determinant is still 12, regardless of translation. Therefore, determinant is translation invariant? Wait, but the determinant is 12 in this case. So even when translated, determinant remains same. So that's interesting. So determinant is invariant under translation. Therefore, no matter how we translate the triangle, determinant remains same. Therefore, even if we rotate the triangle, would determinant change? Let me check with rotation.

Suppose we rotate the triangle. Let me take point A(0,0), B(3,0), C(0,4). Rotate by 90 degrees around origin. Then new points: A'(0,0), B'(0,3), C'(-4,0). Compute determinant:

[0  0  1]

[0  3  1]

[-4 0 1]

Determinant:

0*(3*1 - 0*1) - 0*(0*1 - (-4)*1) + 1*(0*0 - (-4)*3) = 0 - 0 + 1*(0 + 12) = 12. Same determinant. So determinant is invariant under rotation. Therefore, determinant is same regardless of rotation and translation. Therefore, regardless of coordinate system, determinant is 12, so squared is 144. Therefore, answer is 144.

Therefore, regardless of the positions of the points, as long as they form a triangle with sides 3,4,5, determinant is 12, square is 144. Therefore, the answer is 144.

But let me check if there's a case where the determinant could be different. Suppose the triangle is not right-angled. Wait, but in our case, the distances are 3,4,5, which is fixed. So regardless of how the triangle is placed, the area is fixed. Therefore, determinant is fixed. Therefore, answer is 144.

But maybe to make sure, let me think of another configuration. Suppose points are not placed in the coordinate system as I did. Let me assign coordinates differently. Suppose I have points A(0,0), B(0,3), C(4,0). Wait, this is same as before, but swapping coordinates. Then the determinant would be:

[0  0  1]

[0  3  1]

[4  0  1]

Determinant:

0*(3*1 - 0*1) - 0*(0*1 - 4*1) + 1*(0*0 - 4*3) = 0 - 0 + 1*(0 - 12) = -12. Absolute value 12, square is 144. Same result.

Alternatively, suppose I have different coordinates where the triangle is placed in a different way. Let me say point A(1,1), B(1,4), C(5,1). Let me check distances:

AB: sqrt( (1-1)^2 + (4-1)^2 ) = 3, good.

AC: sqrt( (5-1)^2 + (1-1)^2 ) = 4, good.

BC: sqrt( (5-1)^2 + (1-4)^2 ) = sqrt(16 + 9) = 5, good.

Now compute determinant:

[1  1  1]

[1  4  1]

[5  1  1]

Determinant calculation:

1*(4*1 - 1*1) - 1*(1*1 - 5*1) + 1*(1*1 - 5*4)

= 1*(4 - 1) - 1*(1 - 5) + 1*(1 - 20)

= 1*3 - 1*(-4) + 1*(-19)

= 3 + 4 -19 = -12. Absolute value 12, square 144. Same result.

Therefore, regardless of coordinates, determinant squared is 144. Therefore, answer is 144.

But let me think again if there is a possibility that the triangle is not right-angled. Wait, given AB² + AC² = BC², so it must be right-angled. Therefore, area is fixed. Therefore determinant squared is fixed. Hence, the answer is 144. Therefore, I can confidently say that the answer is 144.

Wait, but let me think again if there's another approach, maybe using coordinates algebraically. Let me try to compute the determinant squared using the given distances. Maybe using the formula for determinant squared in terms of the squares of distances?

Alternatively, I can recall that the square of the determinant is related to the area squared, which is related to the sides. But since we have the sides, maybe using Heron's formula? Let me try that.

For triangle with sides a=3, b=4, c=5. Area is 6, as we know. So Heron's formula gives:

s = (a + b + c)/2 = (3+4+5)/2 = 6.

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. Correct. Therefore, area is 6. Therefore, determinant squared is (2 * area)^2 = (12)^2 = 144. Same result.

Alternatively, perhaps we can relate the determinant squared with the given squared distances. But seems more complicated. However, just to explore, maybe using coordinates.

Let me denote the determinant as D. Then D^2 = (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))^2.

But maybe expanding this expression? But perhaps complicated. However, given that we already have a solid answer through area, perhaps not necessary.

Alternatively, think of the determinant squared as the square of the area times 4. Since D = 2 * area, so D² = 4 * area². Since area is 6, then 4*36 = 144.

Therefore, answer is 144.

Therefore, I think the answer is 144, so boxed answer is \boxed{144}.

**Final Answer**
\boxed{144}

To solve the problem, we are given three equations representing 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 distances correspond to the lengths of the sides of a triangle with points $ A(x_1, y_1) $, $ B(x_2, y_2) $, and $ C(x_3, y_3) $, denoted as:

- $ AB = 3 $,
- $ AC = 4 $,
- $ BC = 5 $.

We observe that:

$$
AB^2 + AC^2 = 9 + 16 = 25 = BC^2
$$

This confirms that triangle $ ABC $ is a **right triangle** with a right angle at $ A $.

---

### Area of the Triangle

For a right triangle with legs $ AB = 3 $ and $ AC = 4 $, the area is given by:

$$
\text{Area} = \frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} \cdot 3 \cdot 4 = 6
$$

However, the determinant we are asked to consider is the determinant of the following matrix:

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

This determinant, when expanded, represents **twice the signed area** of the triangle $ ABC $ (up to a sign). So,

$$
\left| \begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix} \right| = 2 \times \text{Area} = 2 \times 6 = 12
$$

Therefore, the **square of the absolute value of the determinant** is:

$$
\left(\left| \begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix} \right| \right)^2 = 12^2 = \boxed{144}
$$

---

### 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 have this geometry problem here. Let me try to visualize it first. There's a triangle ABC with medians AD, BE, and CF intersecting at the centroid G. Then there's a line through G that's parallel to BC, and this line intersects AB and AC at points M and N, respectively. The area of triangle ABC is given as 144, and I need to find the area of triangle ENG. Wait, hold on, the problem mentions triangle ENG, but I don't remember where point E is. Wait, let me check again.

The problem says: "The medians AD, BE, and CF of triangle ABC intersect at the centroid G. The line through G that is parallel to BC intersects AB and AC at M and N, respectively. If the area of triangle ABC is 144, then find the area of triangle ENG."

Wait, so point E is one of the medians? Since BE is a median, E must be the midpoint of AC? Wait, no. Wait, the medians are AD, BE, and CF. So, AD is a median from A to BC, so D is the midpoint of BC. Similarly, BE is a median from B to AC, so E is the midpoint of AC. And CF is a median from C to AB, so F is the midpoint of AB. Got it. So E is the midpoint of AC. So then ENG is triangle formed by points E, N, and G. So we need to find the area of triangle ENG.

Alright, let me try to draw this triangle in my mind. Triangle ABC with centroid G. The line through G parallel to BC cuts AB at M and AC at N. So that line MN is parallel to BC and passes through the centroid. Then, E is the midpoint of AC. So we have points E, N, G. Need to find area of triangle ENG.

First, maybe I can assign coordinates to the triangle ABC to make this easier. Coordinates might help. Let me try coordinate geometry.

Let me place triangle ABC such that calculations are easier. Let me let BC be horizontal for simplicity. Let me assign coordinates:

Let me suppose that point B is at (0, 0), point C is at (2c, 0), so that BC has length 2c, making the midpoint D at (c, 0). Then, let me let point A be at (a, b). Then, since E is the midpoint of AC, coordinates of E would be the average of coordinates of A and C. So E_x = (a + 2c)/2, E_y = (b + 0)/2 = b/2. Similarly, F is the midpoint of AB, so coordinates of F would be ((a + 0)/2, (b + 0)/2) = (a/2, b/2). Similarly, D is midpoint of BC: (c, 0).

The centroid G is the intersection of the medians. The centroid divides each median in the ratio 2:1. So coordinates of centroid G can be found by averaging coordinates of A, B, C. So G_x = (a + 0 + 2c)/3, G_y = (b + 0 + 0)/3 = b/3.

Alternatively, since centroid divides the median AD in 2:1, so starting from A, moving towards D, the centroid is 2/3 of the way. But coordinates would be same as average of A, B, C.

But maybe using coordinates is going to get messy. Let me see if I can find ratios instead.

Alternatively, maybe using vectors? But maybe coordinates is okay.

Alternatively, since we know the area of ABC is 144, perhaps we can find ratios of areas based on similar triangles and properties of centroid.

First, since MN is parallel to BC and passes through centroid G. Let me recall that in a triangle, a line parallel to the base divides the sides proportionally. Also, centroid divides the medians in 2:1.

Let me think about the line MN. Since MN is parallel to BC, triangle AMN is similar to triangle ABC. Let me denote the ratio of similarity as k. Then the area of triangle AMN would be k² times area of ABC. However, I need to find where exactly MN is located. Since it passes through centroid G.

But how can I find k? Let me think.

Let me consider the coordinates approach again. Let me assign coordinates to ABC as follows:

Let me take coordinates for simplicity. Let me let ABC be such that coordinates are:

Let me take coordinates where B is (0, 0), C is (2, 0), and A is (0, 2). Wait, but then area of ABC would be (base * height)/2 = (2*2)/2 = 2. But area is given as 144. Maybe I can later scale up. Alternatively, maybe assign coordinates more generally.

Alternatively, let me take coordinates such that BC is horizontal, and centroid G is at (something). Wait, but maybe coordinates can be chosen so that calculations are easier.

Alternatively, since MN is parallel to BC and passes through centroid G, maybe I can find the ratio in which G divides MN? Wait, but MN is a line through G parallel to BC. So, if I can find the position of MN relative to BC, then I can find coordinates of M and N.

Alternatively, think in terms of vectors. Let me try that.

Let me denote vectors with position vectors from origin. Let me set coordinate system with G as origin? Maybe not. Let me try another approach.

Let me recall that the centroid divides the medians in ratio 2:1. So, for example, in median BE, which goes from B to E (midpoint of AC), centroid G divides BE such that BG:GE = 2:1. Similarly for other medians.

Now, the line MN is parallel to BC and passes through G. Let me think about how to find MN.

Since MN is parallel to BC, then triangle AMN ~ triangle ABC. Let me find the ratio of similarity.

Let me denote the ratio as k. Then, since MN is parallel to BC, the distance from A to MN is k times the distance from A to BC. But since G is on MN, maybe I can relate the distances.

Alternatively, since centroid is located at 1/3 of the height from the base. Wait, the centroid is located at 1/3 the distance from the base. Wait, if we consider the median from A to BC, the centroid is 2/3 along the median from A towards BC.

But MN is a line parallel to BC passing through G. So, if I think of MN as a line parallel to BC, then the distance between MN and BC would be... Let me think in terms of heights.

Let me suppose that the height of triangle ABC from A to BC is h. Then the centroid is located at 2/3 of the height from BC, so the distance from centroid G to BC is 2/3 h? Wait, no. Wait, the centroid is located 2/3 along the median from the vertex. So if the median from A to BC has length, say, m, then centroid divides it into 2/3 m from A and 1/3 m from BC. Therefore, the distance from centroid to BC is 1/3 of the height h.

Therefore, if MN is parallel to BC and passes through G, then the distance from A to MN is h - 1/3 h = 2/3 h. Therefore, the ratio of similarity k between triangle AMN and ABC is 2/3? Wait, because the height is scaled by 2/3? Wait, but if the distance from A to MN is 2/3 h, then the ratio of similarity is 2/3? Wait, no. Wait, if the distance from A to BC is h, then the distance from A to MN is 2/3 h? Wait, no. Wait, the centroid is located 1/3 h above BC? Wait, no:

Wait, when you have a median from A to BC, the centroid is located 2/3 along the median from A. Therefore, the distance from A to centroid is 2/3 of the median's length, but the distance from centroid to BC is 1/3 of the median's length. However, the height h of ABC is the length of the altitude from A to BC. So the length of the median from A to BC is not necessarily equal to the altitude unless the triangle is isoceles. Therefore, maybe my previous thought is not accurate. Maybe I need to think differently.

Alternatively, maybe using coordinate geometry again. Let me try coordinates again with a specific triangle.

Let me assume coordinates where BC is on x-axis, B at (0,0), C at (2, 0), A at (0, 2). Then ABC is a right triangle with legs of length 2, area is (2*2)/2 = 2. Then centroid G is at ((0 + 0 + 2)/3, (2 + 0 + 0)/3) = (2/3, 2/3). Now, the line through G parallel to BC (which is horizontal) is horizontal line y = 2/3. This line intersects AB and AC at M and N.

First, find equations of AB and AC.

Equation of AB: from (0, 2) to (0, 0). Wait, AB is vertical line x=0. Wait, but if A is (0, 2), B is (0, 0), then AB is vertical. But then a horizontal line y = 2/3 would intersect AB at (0, 2/3). Similarly, equation of AC: from A(0,2) to C(2, 0). The slope is (0 - 2)/(2 - 0) = -1. So equation is y - 2 = -1(x - 0) => y = -x + 2. The horizontal line y = 2/3 intersects AC where 2/3 = -x + 2 => x = 2 - 2/3 = 4/3. Therefore, point N is at (4/3, 2/3). So M is at (0, 2/3). Thus, MN is the line segment from (0, 2/3) to (4/3, 2/3). Now, need to find area of triangle ENG. Wait, E is the midpoint of AC. Coordinates of E: midpoint of AC: A(0, 2) and C(2, 0), so midpoint E is ((0 + 2)/2, (2 + 0)/2) = (1, 1). So E is (1, 1). Point G is (2/3, 2/3). Point N is (4/3, 2/3). So triangle ENG has coordinates E(1,1), N(4/3, 2/3), G(2/3, 2/3). Let me compute the area of triangle ENG.

First, let me write coordinates:

E: (1, 1)

N: (4/3, 2/3)

G: (2/3, 2/3)

We can use the shoelace formula.

Compute area as ½ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Plugging in:

x1 = 1, y1 = 1

x2 = 4/3, y2 = 2/3

x3 = 2/3, y3 = 2/3

Compute:

1*(2/3 - 2/3) + (4/3)*(2/3 - 1) + (2/3)*(1 - 2/3)

Simplify:

First term: 1*(0) = 0

Second term: (4/3)*( -1/3) = -4/9

Third term: (2/3)*(1/3) = 2/9

Total: 0 - 4/9 + 2/9 = (-2)/9

Take absolute value and multiply by 1/2: ½ * | -2/9 | = ½ * 2/9 = 1/9.

So area of triangle ENG in this coordinate system is 1/9. But the area of triangle ABC in this case is 2. The problem says area ABC is 144, so need to scale up.

Since in this coordinate system, area ABC is 2, and we need it to be 144. The scaling factor squared times 2 equals 144. So scaling factor k satisfies k² * 2 = 144 => k² = 72 => k = sqrt(72). Therefore, areas scale by k², so the area ENG in the actual problem would be (1/9) * (144 / 2) = (1/9)*72 = 8. Wait, is that correct? Wait, if original area is 2 and we want it to be 144, then scaling factor squared is 144 / 2 = 72, so scaling factor is sqrt(72). Then areas of figures scale by that factor squared. Wait, but the area of ENG in the original coordinate system is 1/9. So scaling up by 72 would make it (1/9) * 72 = 8. Therefore, the answer is 8?

Wait, but let me verify this with another approach to make sure.

Alternatively, maybe my coordinate system was too specific? Let me check with another coordinate system to confirm.

Alternatively, perhaps I made an error in the coordinate system. Let me think again.

Wait, in my coordinate system, ABC has coordinates A(0, 2), B(0, 0), C(2, 0). Then area is 2, and the area of ENG is 1/9. Therefore, ratio of areas is 1/9 divided by 2 is 1/18. Therefore, if ABC is 144, then ENG is 144 * (1/18) = 8. So that seems correct. But let me confirm with another coordinate system.

Alternatively, take a different triangle. Let me take ABC with coordinates A(0, 0), B(0, 0)? No, that would collapse. Let me pick a general triangle. Let me take coordinates where ABC is any triangle with centroid at G, and compute the area ratio.

Alternatively, maybe think in terms of mass point or ratios.

Since MN is parallel to BC, then AM/AB = AN/AC = k. Let me find k. Since MN passes through centroid G. Let me think of the position of G.

In triangle ABC, centroid divides medians in ratio 2:1. Let me consider median BE. E is midpoint of AC. Let me parametrize the line BE. Let me suppose that point G is located 2/3 of the way from B to E. Now, the line MN is parallel to BC and passes through G. Let me find the ratio k such that AM/AB = AN/AC = k.

Since MN is parallel to BC, then by basic proportionality theorem (Thales' theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those sides proportionally. Therefore, AM/AB = AN/AC = k. Let me find k such that the line MN passes through centroid G.

Let me express coordinates again in terms of vectors or ratios. Let me use barycentric coordinates or parametric equations.

Alternatively, think of coordinates again but more generally. Let me consider triangle ABC with coordinates:

Let me let A be (0, 0), B be (b, 0), and C be (c, h). Then centroid G is at ((b + c)/3, h/3). The line MN is parallel to BC. The coordinates of BC are from (b, 0) to (c, h). So the direction vector of BC is (c - b, h). Therefore, the line MN is parallel to BC, so it has the same direction vector. However, since MN passes through G, we need to find points M on AB and N on AC such that MN is parallel to BC and passes through G.

Alternatively, since MN is parallel to BC, the coordinates of points M and N can be parameterized.

Let me parametrize AB. AB goes from A(0,0) to B(b, 0). Let me let point M divide AB in ratio t:1-t. Wait, let me use parameter k such that AM = k AB. So coordinates of M would be (kb, 0). Similarly, since MN is parallel to BC, point N on AC should be such that AN = k AC. Coordinates of AC go from A(0,0) to C(c, h). So coordinates of N would be (kc, kh). Then the line MN connects (kb, 0) and (kc, kh). The vector MN is (kc - kb, kh - 0) = (k(c - b), kh). Which is the same direction as BC vector (c - b, h). So this is correct, since MN is parallel to BC.

Now, we need this line MN to pass through centroid G. Coordinates of G are ((b + c)/3, h/3). Therefore, the point G lies on MN. Let me parametrize MN. Let me write parametric equations for MN. Let me use parameter s.

Starting at M(kb, 0), direction vector (k(c - b), kh). So any point on MN can be written as (kb + s * k(c - b), 0 + s * kh). We need this to pass through G(( (b + c)/3, h/3 )) for some s.

Therefore, set up equations:

kb + s * k(c - b) = (b + c)/3

and

s * kh = h/3

From the second equation: s * kh = h/3 => s = 1/(3k) (assuming h ≠ 0, which it is not in a triangle).

Substitute s into first equation:

kb + (1/(3k)) * k(c - b) = (b + c)/3

Simplify:

kb + (c - b)/3 = (b + c)/3

Multiply through by 3 to eliminate denominators:

3kb + (c - b) = b + c

Simplify left side:

3kb + c - b = b + c

Subtract c from both sides:

3kb - b = b

Factor left side:

b(3k - 1) = b

Assuming b ≠ 0 (since AB is a side of triangle, so b ≠ 0):

3k - 1 = 1 => 3k = 2 => k = 2/3.

Therefore, the ratio k is 2/3. Therefore, AM = (2/3) AB, AN = (2/3) AC. Therefore, MN divides AB and AC in ratio 2:1, so that the line MN is parallel to BC and passes through centroid G.

So in this general case, with coordinates as above, MN is located such that AM = (2/3) AB, AN = (2/3) AC. Therefore, triangle AMN is similar to triangle ABC with ratio 2/3, so area ratio is (2/3)^2 = 4/9. Therefore, area of AMN is 4/9 area ABC. But since area ABC is 144, area AMN would be 4/9 * 144 = 64. But the problem is not asking for AMN, but for ENG.

So need to find area of triangle ENG. Let me find coordinates of E, N, G.

Point E is midpoint of AC. Coordinates of A(0,0), C(c, h), so E is at ((c)/2, h/2). Coordinates of N: as above, N is (kc, kh) = ( (2/3)c, (2/3)h ). Coordinates of G: ((b + c)/3, h/3). Therefore, E is (c/2, h/2), N is (2c/3, 2h/3), G is ((b + c)/3, h/3). Now need to find area of triangle ENG.

Let me use the shoelace formula again with these coordinates.

Coordinates:

E: (c/2, h/2)

N: (2c/3, 2h/3)

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

Let me write them as:

E: (x1, y1) = (c/2, h/2)

N: (x2, y2) = (2c/3, 2h/3)

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

Compute area using formula:

Area = ½ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Let me compute each term:

First term: x1(y2 - y3) = (c/2)(2h/3 - h/3) = (c/2)(h/3) = ch/6

Second term: x2(y3 - y1) = (2c/3)(h/3 - h/2) = (2c/3)( (2h - 3h)/6 ) = (2c/3)( -h/6 ) = -2c h / 18 = -c h / 9

Third term: x3(y1 - y2) = ((b + c)/3)(h/2 - 2h/3) = ((b + c)/3)( (3h - 4h)/6 ) = ((b + c)/3)( -h/6 ) = - (b + c) h / 18

Adding these three terms:

ch/6 - ch/9 - (b + c)h / 18

Let me convert all terms to eighteenths:

ch/6 = 3ch/18

- ch/9 = -2ch/18

- (b + c)h /18 remains as is.

So total:

(3ch - 2ch - (b + c)h ) / 18 = (ch - bh - ch)/18 = (-bh)/18

Take absolute value and multiply by ½:

Area = ½ * | -bh / 18 | = ½ * bh / 18 = bh / 36

So area of triangle ENG is bh / 36.

But in our coordinate system, the area of triangle ABC is given by base * height / 2. Let me check that. In this coordinate system, ABC has coordinates A(0,0), B(b, 0), C(c, h). Wait, actually, if we take AB as base, but the coordinates are not necessarily right triangle. Wait, to compute area of ABC, using coordinates, we can use determinant formula:

Area ABC = ½ | (Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By)) | = ½ | 0*(0 - h) + b*(h - 0) + c*(0 - 0) | = ½ | 0 + bh + 0 | = bh / 2.

So area ABC = bh / 2. Therefore, the area of ENG is (bh / 36) = ( (bh / 2 ) ) * (1/18 ) = area ABC * 1/18.

Therefore, area ENG = (1/18) * area ABC.

Since area ABC is 144, then area ENG = 144 / 18 = 8. So answer is 8. Which matches with my previous coordinate example where with area ABC = 2, area ENG was 1/9, and 2*(1/18) = 1/9, so yes, 1/18 ratio. Therefore, answer is 8.

Therefore, the answer is \boxed{8}.

But let me check once again with another approach to make sure.

Alternative approach:

Since we found that in general, area ENG is 1/18 of area ABC. So if area ABC is 144, then area ENG is 144 / 18 = 8. So answer is 8. Seems solid.

Alternatively, maybe use vectors.

Let me consider vectors. Let me take A as origin. So position vectors:

Let me denote vectors:

Let vector AB = **b**, vector AC = **c**. Then coordinates:

Point A: 0

Point B: **b**

Point C: **c**

Centroid G: ( **b** + **c** ) / 3 + 0? Wait, centroid is (A + B + C)/3. Since A is origin, it's ( **b** + **c** ) / 3.

Point E is midpoint of AC: (0 + **c**)/2 = **c**/2.

Point N is on AC such that AN = (2/3) AC. Since AC is from A(0) to C(**c**), so N = (2/3)**c**.

Similarly, point M is on AB such that AM = (2/3) AB = (2/3)**b**.

Line MN goes from (2/3)**b** to (2/3)**c**.

We need to find area of triangle ENG, where E is **c**/2, N is (2/3)**c**, G is ( **b** + **c** ) / 3.

So points E, N, G have position vectors:

E: **c**/2

N: (2/3)**c**

G: ( **b** + **c** ) / 3

To compute area of triangle ENG, we can use cross product formula.

Area = (1/2) | (N - E) × (G - E) |.

Let me compute vectors N - E and G - E.

First, N - E = (2/3 **c**) - (1/2 **c**) = (4/6 - 3/6)**c** = (1/6)**c**

G - E = [ ( **b** + **c** ) / 3 ] - (1/2 **c**) = ( **b** / 3 + **c** / 3 - **c** / 2 ) = **b** / 3 + ( 2**c** / 6 - 3**c** / 6 ) = **b** / 3 - **c** / 6

Now, cross product (N - E) × (G - E) = (1/6 **c**) × ( **b** / 3 - **c** / 6 ) = (1/6)**c** × **b** / 3 - (1/6)**c** × **c** / 6

But cross product of any vector with itself is zero, so second term is 0. Therefore:

= (1/6)(1/3) **c** × **b** = (1/18)( **c** × **b** )

The magnitude of this cross product is (1/18)|**c** × **b**|.

Therefore, area of triangle ENG is (1/2)(1/18)|**c** × **b**| = (1/36)|**c** × **b**|.

But area of triangle ABC is (1/2)|**b** × **c**|, so area ENG is (1/36)|**c** × **b**| = (1/18)(1/2)|**b** × **c**| = (1/18) area ABC. Therefore, same result. Therefore, area ENG = (1/18)*144=8. So answer is 8. Definitely.

Therefore, regardless of coordinate system or approach, answer is 8. So I think that's correct.

**Final Answer**
\boxed{8}

To solve the problem, we begin by analyzing the geometry and coordinate relationships within triangle $ ABC $, with the centroid $ G $, and the line through $ G $ parallel to side $ BC $ intersecting sides $ AB $ and $ AC $ at points $ M $ and $ N $, respectively. The goal is to find the area of triangle $ ENG $, where $ E $ is the midpoint of side $ AC $, and the area of triangle $ ABC $ is given as 144.

---

### Step 1: Coordinate System

We assign coordinates to triangle $ ABC $ to simplify the computation:

- Let $ A = (0, 0) $, $ B = (b, 0) $, and $ C = (c, h) $. This sets up a general triangle with base $ AB $ along the x-axis.

- Then, the midpoint $ E $ of $ AC $ is:
  $$
  E = \left( \frac{c}{2}, \frac{h}{2} \right)
  $$

- The centroid $ G $ of triangle $ ABC $ is the average of the coordinates of the vertices:
  $$
  G = \left( \frac{b + c}{3}, \frac{h}{3} \right)
  $$

- Since the line through $ G $ is parallel to $ BC $, and intersects sides $ AB $ and $ AC $, the points $ M $ and $ N $ divide the triangle in a specific ratio along these sides. By the properties of similar triangles and the centroid, we can deduce that this line cuts $ AB $ and $ AC $ at $ \frac{2}{3} $ of the total length from the vertices, i.e.,
  $$
  M = \left( \frac{2b}{3}, 0 \right), \quad N = \left( \frac{2c}{3}, \frac{2h}{3} \right)
  $$

---

### Step 2: Use of Area Formula

Now, we compute the area of triangle $ ENG $ using vector cross product, which is well-suited to this setup.

Let the vectors be:
- $\vec{E} = \left( \frac{c}{2}, \frac{h}{2} \right)$
- $\vec{N} = \left( \frac{2c}{3}, \frac{2h}{3} \right)$
- $\vec{G} = \left( \frac{b + c}{3}, \frac{h}{3} \right)$

Vectors used for area computation:
- $\vec{N} - \vec{E} = \left( \frac{2c}{3} - \frac{c}{2}, \frac{2h}{3} - \frac{h}{2} \right) = \left( \frac{c}{6}, \frac{h}{6} \right)$
- $\vec{G} - \vec{E} = \left( \frac{b}{3} - \frac{c}{2}, \frac{h}{3} - \frac{h}{2} \right) = \left( \frac{2b - 3c}{6}, -\frac{h}{6} \right)$

Now compute the magnitude of the cross product (in 2D, we take the determinant):

$$
\text{Area} = \frac{1}{2} \left| \vec{N} - \vec{E} \right| \times \left| \vec{G} - \vec{E} \right|
$$

But instead of computing determinants directly, we observe a more elegant pattern: all computations show that the area of triangle $ ENG $ is **1/18** of the area of triangle $ ABC $.

---

### Step 3: Final Calculation

Since the area of triangle $ ABC $ is given as 144, the area of triangle $ ENG $ is:

$$
\text{Area of } ENG = \frac{1}{18} \times 144 = \boxed{8}
$$

---

### Final Answer
$$
\boxed{8}
$$

[Assistant]
Explanation:
Okay, so I have this geometry problem here. Let me try to visualize it first. There's a triangle ABC with medians AD, BE, and CF intersecting at the centroid G. Then there's a line through G that's parallel to BC, and this line intersects AB and AC at points M and N, respectively. The area of triangle ABC is given as 144, and I need to find the area of triangle ENG. Wait, hold on, the problem mentions triangle ENG, but I don't remember where point E is. Wait, let me check again.

The problem says: "The medians AD, BE, and CF of triangle ABC intersect at the centroid G. The line through G that is parallel to BC intersects AB and AC at M and N, respectively. If the area of triangle ABC is 144, then find the area of triangle ENG."

Wait, so point E is one of the medians? Since BE is a median, E must be the midpoint of AC? Wait, no. Wait, the medians are AD, BE, and CF. So, AD is a median from A to BC, so D is the midpoint of BC. Similarly, BE is a median from B to AC, so E is the midpoint of AC. And CF is a median from C to AB, so F is the midpoint of AB. Got it. So E is the midpoint of AC. So then ENG is triangle formed by points E, N, and G. So we need to find the area of triangle ENG.

Alright, let me try to draw this triangle in my mind. Triangle ABC with centroid G. The line through G parallel to BC cuts AB at M and AC at N. So that line MN is parallel to BC and passes through the centroid. Then, E is the midpoint of AC. So we have points E, N, G. Need to find area of triangle ENG.

First, maybe I can assign coordinates to the triangle ABC to make this easier. Coordinates might help. Let me try coordinate geometry.

Let me place triangle ABC such that calculations are easier. Let me let BC be horizontal for simplicity. Let me assign coordinates:

Let me suppose that point B is at (0, 0), point C is at (2c, 0), so that BC has length 2c, making the midpoint D at (c, 0). Then, let me let point A be at (a, b). Then, since E is the midpoint of AC, coordinates of E would be the average of coordinates of A and C. So E_x = (a + 2c)/2, E_y = (b + 0)/2 = b/2. Similarly, F is the midpoint of AB, so coordinates of F would be ((a + 0)/2, (b + 0)/2) = (a/2, b/2). Similarly, D is midpoint of BC: (c, 0).

The centroid G is the intersection of the medians. The centroid divides each median in the ratio 2:1. So coordinates of centroid G can be found by averaging coordinates of A, B, C. So G_x = (a + 0 + 2c)/3, G_y = (b + 0 + 0)/3 = b/3.

Alternatively, since centroid divides the median AD in 2:1, so starting from A, moving towards D, the centroid is 2/3 of the way. But coordinates would be same as average of A, B, C.

But maybe using coordinates is going to get messy. Let me see if I can find ratios instead.

Alternatively, maybe using vectors? But maybe coordinates is okay.

Alternatively, since we know the area of ABC is 144, perhaps we can find ratios of areas based on similar triangles and properties of centroid.

First, since MN is parallel to BC and passes through centroid G. Let me recall that in a triangle, a line parallel to the base divides the sides proportionally. Also, centroid divides the medians in 2:1.

Let me think about the line MN. Since MN is parallel to BC, triangle AMN is similar to triangle ABC. Let me denote the ratio of similarity as k. Then the area of triangle AMN would be k² times area of ABC. However, I need to find where exactly MN is located. Since it passes through centroid G.

But how can I find k? Let me think.

Let me consider the coordinates approach again. Let me assign coordinates to ABC as follows:

Let me take coordinates for simplicity. Let me let ABC be such that coordinates are:

Let me take coordinates where B is (0, 0), C is (2, 0), and A is (0, 2). Wait, but then area of ABC would be (base * height)/2 = (2*2)/2 = 2. But area is given as 144. Maybe I can later scale up. Alternatively, maybe assign coordinates more generally.

Alternatively, let me take coordinates such that BC is horizontal, and centroid G is at (something). Wait, but maybe coordinates can be chosen so that calculations are easier.

Alternatively, since MN is parallel to BC and passes through centroid G, maybe I can find the ratio in which G divides MN? Wait, but MN is a line through G parallel to BC. So, if I can find the position of MN relative to BC, then I can find coordinates of M and N.

Alternatively, think in terms of vectors. Let me try that.

Let me denote vectors with position vectors from origin. Let me set coordinate system with G as origin? Maybe not. Let me try another approach.

Let me recall that the centroid divides the medians in ratio 2:1. So, for example, in median BE, which goes from B to E (midpoint of AC), centroid G divides BE such that BG:GE = 2:1. Similarly for other medians.

Now, the line MN is parallel to BC and passes through G. Let me think about how to find MN.

Since MN is parallel to BC, then triangle AMN ~ triangle ABC. Let me find the ratio of similarity.

Let me denote the ratio as k. Then, since MN is parallel to BC, the distance from A to MN is k times the distance from A to BC. But since G is on MN, maybe I can relate the distances.

Alternatively, since centroid is located at 1/3 of the height from the base. Wait, the centroid is located at 1/3 the distance from the base. Wait, if we consider the median from A to BC, the centroid is 2/3 along the median from A towards BC.

But MN is a line parallel to BC passing through G. So, if I think of MN as a line parallel to BC, then the distance between MN and BC would be... Let me think in terms of heights.

Let me suppose that the height of triangle ABC from A to BC is h. Then the centroid is located at 2/3 of the height from BC, so the distance from centroid G to BC is 2/3 h? Wait, no. Wait, the centroid is located 2/3 along the median from the vertex. So if the median from A to BC has length, say, m, then centroid divides it into 2/3 m from A and 1/3 m from BC. Therefore, the distance from centroid to BC is 1/3 of the height h.

Therefore, if MN is parallel to BC and passes through G, then the distance from A to MN is h - 1/3 h = 2/3 h. Therefore, the ratio of similarity k between triangle AMN and ABC is 2/3? Wait, because the height is scaled by 2/3? Wait, but if the distance from A to MN is 2/3 h, then the ratio of similarity is 2/3? Wait, no. Wait, if the distance from A to BC is h, then the distance from A to MN is 2/3 h? Wait, no. Wait, the centroid is located 1/3 h above BC? Wait, no:

Wait, when you have a median from A to BC, the centroid is located 2/3 along the median from A. Therefore, the distance from A to centroid is 2/3 of the median's length, but the distance from centroid to BC is 1/3 of the median's length. However, the height h of ABC is the length of the altitude from A to BC. So the length of the median from A to BC is not necessarily equal to the altitude unless the triangle is isoceles. Therefore, maybe my previous thought is not accurate. Maybe I need to think differently.

Alternatively, maybe using coordinate geometry again. Let me try coordinates again with a specific triangle.

Let me assume coordinates where BC is on x-axis, B at (0,0), C at (2, 0), A at (0, 2). Then ABC is a right triangle with legs of length 2, area is (2*2)/2 = 2. Then centroid G is at ((0 + 0 + 2)/3, (2 + 0 + 0)/3) = (2/3, 2/3). Now, the line through G parallel to BC (which is horizontal) is horizontal line y = 2/3. This line intersects AB and AC at M and N.

First, find equations of AB and AC.

Equation of AB: from (0, 2) to (0, 0). Wait, AB is vertical line x=0. Wait, but if A is (0, 2), B is (0, 0), then AB is vertical. But then a horizontal line y = 2/3 would intersect AB at (0, 2/3). Similarly, equation of AC: from A(0,2) to C(2, 0). The slope is (0 - 2)/(2 - 0) = -1. So equation is y - 2 = -1(x - 0) => y = -x + 2. The horizontal line y = 2/3 intersects AC where 2/3 = -x + 2 => x = 2 - 2/3 = 4/3. Therefore, point N is at (4/3, 2/3). So M is at (0, 2/3). Thus, MN is the line segment from (0, 2/3) to (4/3, 2/3). Now, need to find area of triangle ENG. Wait, E is the midpoint of AC. Coordinates of E: midpoint of AC: A(0, 2) and C(2, 0), so midpoint E is ((0 + 2)/2, (2 + 0)/2) = (1, 1). So E is (1, 1). Point G is (2/3, 2/3). Point N is (4/3, 2/3). So triangle ENG has coordinates E(1,1), N(4/3, 2/3), G(2/3, 2/3). Let me compute the area of triangle ENG.

First, let me write coordinates:

E: (1, 1)

N: (4/3, 2/3)

G: (2/3, 2/3)

We can use the shoelace formula.

Compute area as ½ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Plugging in:

x1 = 1, y1 = 1

x2 = 4/3, y2 = 2/3

x3 = 2/3, y3 = 2/3

Compute:

1*(2/3 - 2/3) + (4/3)*(2/3 - 1) + (2/3)*(1 - 2/3)

Simplify:

First term: 1*(0) = 0

Second term: (4/3)*( -1/3) = -4/9

Third term: (2/3)*(1/3) = 2/9

Total: 0 - 4/9 + 2/9 = (-2)/9

Take absolute value and multiply by 1/2: ½ * | -2/9 | = ½ * 2/9 = 1/9.

So area of triangle ENG in this coordinate system is 1/9. But the area of triangle ABC in this case is 2. The problem says area ABC is 144, so need to scale up.

Since in this coordinate system, area ABC is 2, and we need it to be 144. The scaling factor squared times 2 equals 144. So scaling factor k satisfies k² * 2 = 144 => k² = 72 => k = sqrt(72). Therefore, areas scale by k², so the area ENG in the actual problem would be (1/9) * (144 / 2) = (1/9)*72 = 8. Wait, is that correct? Wait, if original area is 2 and we want it to be 144, then scaling factor squared is 144 / 2 = 72, so scaling factor is sqrt(72). Then areas of figures scale by that factor squared. Wait, but the area of ENG in the original coordinate system is 1/9. So scaling up by 72 would make it (1/9) * 72 = 8. Therefore, the answer is 8?

Wait, but let me verify this with another approach to make sure.

Alternatively, maybe my coordinate system was too specific? Let me check with another coordinate system to confirm.

Alternatively, perhaps I made an error in the coordinate system. Let me think again.

Wait, in my coordinate system, ABC has coordinates A(0, 2), B(0, 0), C(2, 0). Then area is 2, and the area of ENG is 1/9. Therefore, ratio of areas is 1/9 divided by 2 is 1/18. Therefore, if ABC is 144, then ENG is 144 * (1/18) = 8. So that seems correct. But let me confirm with another coordinate system.

Alternatively, take a different triangle. Let me take ABC with coordinates A(0, 0), B(0, 0)? No, that would collapse. Let me pick a general triangle. Let me take coordinates where ABC is any triangle with centroid at G, and compute the area ratio.

Alternatively, maybe think in terms of mass point or ratios.

Since MN is parallel to BC, then AM/AB = AN/AC = k. Let me find k. Since MN passes through centroid G. Let me think of the position of G.

In triangle ABC, centroid divides medians in ratio 2:1. Let me consider median BE. E is midpoint of AC. Let me parametrize the line BE. Let me suppose that point G is located 2/3 of the way from B to E. Now, the line MN is parallel to BC and passes through G. Let me find the ratio k such that AM/AB = AN/AC = k.

Since MN is parallel to BC, then by basic proportionality theorem (Thales' theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those sides proportionally. Therefore, AM/AB = AN/AC = k. Let me find k such that the line MN passes through centroid G.

Let me express coordinates again in terms of vectors or ratios. Let me use barycentric coordinates or parametric equations.

Alternatively, think of coordinates again but more generally. Let me consider triangle ABC with coordinates:

Let me let A be (0, 0), B be (b, 0), and C be (c, h). Then centroid G is at ((b + c)/3, h/3). The line MN is parallel to BC. The coordinates of BC are from (b, 0) to (c, h). So the direction vector of BC is (c - b, h). Therefore, the line MN is parallel to BC, so it has the same direction vector. However, since MN passes through G, we need to find points M on AB and N on AC such that MN is parallel to BC and passes through G.

Alternatively, since MN is parallel to BC, the coordinates of points M and N can be parameterized.

Let me parametrize AB. AB goes from A(0,0) to B(b, 0). Let me let point M divide AB in ratio t:1-t. Wait, let me use parameter k such that AM = k AB. So coordinates of M would be (kb, 0). Similarly, since MN is parallel to BC, point N on AC should be such that AN = k AC. Coordinates of AC go from A(0,0) to C(c, h). So coordinates of N would be (kc, kh). Then the line MN connects (kb, 0) and (kc, kh). The vector MN is (kc - kb, kh - 0) = (k(c - b), kh). Which is the same direction as BC vector (c - b, h). So this is correct, since MN is parallel to BC.

Now, we need this line MN to pass through centroid G. Coordinates of G are ((b + c)/3, h/3). Therefore, the point G lies on MN. Let me parametrize MN. Let me write parametric equations for MN. Let me use parameter s.

Starting at M(kb, 0), direction vector (k(c - b), kh). So any point on MN can be written as (kb + s * k(c - b), 0 + s * kh). We need this to pass through G(( (b + c)/3, h/3 )) for some s.

Therefore, set up equations:

kb + s * k(c - b) = (b + c)/3

and

s * kh = h/3

From the second equation: s * kh = h/3 => s = 1/(3k) (assuming h ≠ 0, which it is not in a triangle).

Substitute s into first equation:

kb + (1/(3k)) * k(c - b) = (b + c)/3

Simplify:

kb + (c - b)/3 = (b + c)/3

Multiply through by 3 to eliminate denominators:

3kb + (c - b) = b + c

Simplify left side:

3kb + c - b = b + c

Subtract c from both sides:

3kb - b = b

Factor left side:

b(3k - 1) = b

Assuming b ≠ 0 (since AB is a side of triangle, so b ≠ 0):

3k - 1 = 1 => 3k = 2 => k = 2/3.

Therefore, the ratio k is 2/3. Therefore, AM = (2/3) AB, AN = (2/3) AC. Therefore, MN divides AB and AC in ratio 2:1, so that the line MN is parallel to BC and passes through centroid G.

So in this general case, with coordinates as above, MN is located such that AM = (2/3) AB, AN = (2/3) AC. Therefore, triangle AMN is similar to triangle ABC with ratio 2/3, so area ratio is (2/3)^2 = 4/9. Therefore, area of AMN is 4/9 area ABC. But since area ABC is 144, area AMN would be 4/9 * 144 = 64. But the problem is not asking for AMN, but for ENG.

So need to find area of triangle ENG. Let me find coordinates of E, N, G.

Point E is midpoint of AC. Coordinates of A(0,0), C(c, h), so E is at ((c)/2, h/2). Coordinates of N: as above, N is (kc, kh) = ( (2/3)c, (2/3)h ). Coordinates of G: ((b + c)/3, h/3). Therefore, E is (c/2, h/2), N is (2c/3, 2h/3), G is ((b + c)/3, h/3). Now need to find area of triangle ENG.

Let me use the shoelace formula again with these coordinates.

Coordinates:

E: (c/2, h/2)

N: (2c/3, 2h/3)

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

Let me write them as:

E: (x1, y1) = (c/2, h/2)

N: (x2, y2) = (2c/3, 2h/3)

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

Compute area using formula:

Area = ½ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

Let me compute each term:

First term: x1(y2 - y3) = (c/2)(2h/3 - h/3) = (c/2)(h/3) = ch/6

Second term: x2(y3 - y1) = (2c/3)(h/3 - h/2) = (2c/3)( (2h - 3h)/6 ) = (2c/3)( -h/6 ) = -2c h / 18 = -c h / 9

Third term: x3(y1 - y2) = ((b + c)/3)(h/2 - 2h/3) = ((b + c)/3)( (3h - 4h)/6 ) = ((b + c)/3)( -h/6 ) = - (b + c) h / 18

Adding these three terms:

ch/6 - ch/9 - (b + c)h / 18

Let me convert all terms to eighteenths:

ch/6 = 3ch/18

- ch/9 = -2ch/18

- (b + c)h /18 remains as is.

So total:

(3ch - 2ch - (b + c)h ) / 18 = (ch - bh - ch)/18 = (-bh)/18

Take absolute value and multiply by ½:

Area = ½ * | -bh / 18 | = ½ * bh / 18 = bh / 36

So area of triangle ENG is bh / 36.

But in our coordinate system, the area of triangle ABC is given by base * height / 2. Let me check that. In this coordinate system, ABC has coordinates A(0,0), B(b, 0), C(c, h). Wait, actually, if we take AB as base, but the coordinates are not necessarily right triangle. Wait, to compute area of ABC, using coordinates, we can use determinant formula:

Area ABC = ½ | (Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By)) | = ½ | 0*(0 - h) + b*(h - 0) + c*(0 - 0) | = ½ | 0 + bh + 0 | = bh / 2.

So area ABC = bh / 2. Therefore, the area of ENG is (bh / 36) = ( (bh / 2 ) ) * (1/18 ) = area ABC * 1/18.

Therefore, area ENG = (1/18) * area ABC.

Since area ABC is 144, then area ENG = 144 / 18 = 8. So answer is 8. Which matches with my previous coordinate example where with area ABC = 2, area ENG was 1/9, and 2*(1/18) = 1/9, so yes, 1/18 ratio. Therefore, answer is 8.

Therefore, the answer is \boxed{8}.

But let me check once again with another approach to make sure.

Alternative approach:

Since we found that in general, area ENG is 1/18 of area ABC. So if area ABC is 144, then area ENG is 144 / 18 = 8. So answer is 8. Seems solid.

Alternatively, maybe use vectors.

Let me consider vectors. Let me take A as origin. So position vectors:

Let me denote vectors:

Let vector AB = **b**, vector AC = **c**. Then coordinates:

Point A: 0

Point B: **b**

Point C: **c**

Centroid G: ( **b** + **c** ) / 3 + 0? Wait, centroid is (A + B + C)/3. Since A is origin, it's ( **b** + **c** ) / 3.

Point E is midpoint of AC: (0 + **c**)/2 = **c**/2.

Point N is on AC such that AN = (2/3) AC. Since AC is from A(0) to C(**c**), so N = (2/3)**c**.

Similarly, point M is on AB such that AM = (2/3) AB = (2/3)**b**.

Line MN goes from (2/3)**b** to (2/3)**c**.

We need to find area of triangle ENG, where E is **c**/2, N is (2/3)**c**, G is ( **b** + **c** ) / 3.

So points E, N, G have position vectors:

E: **c**/2

N: (2/3)**c**

G: ( **b** + **c** ) / 3

To compute area of triangle ENG, we can use cross product formula.

Area = (1/2) | (N - E) × (G - E) |.

Let me compute vectors N - E and G - E.

First, N - E = (2/3 **c**) - (1/2 **c**) = (4/6 - 3/6)**c** = (1/6)**c**

G - E = [ ( **b** + **c** ) / 3 ] - (1/2 **c**) = ( **b** / 3 + **c** / 3 - **c** / 2 ) = **b** / 3 + ( 2**c** / 6 - 3**c** / 6 ) = **b** / 3 - **c** / 6

Now, cross product (N - E) × (G - E) = (1/6 **c**) × ( **b** / 3 - **c** / 6 ) = (1/6)**c** × **b** / 3 - (1/6)**c** × **c** / 6

But cross product of any vector with itself is zero, so second term is 0. Therefore:

= (1/6)(1/3) **c** × **b** = (1/18)( **c** × **b** )

The magnitude of this cross product is (1/18)|**c** × **b**|.

Therefore, area of triangle ENG is (1/2)(1/18)|**c** × **b**| = (1/36)|**c** × **b**|.

But area of triangle ABC is (1/2)|**b** × **c**|, so area ENG is (1/36)|**c** × **b**| = (1/18)(1/2)|**b** × **c**| = (1/18) area ABC. Therefore, same result. Therefore, area ENG = (1/18)*144=8. So answer is 8. Definitely.

Therefore, regardless of coordinate system or approach, answer is 8. So I think that's correct.

**Final Answer**
\boxed{8}

To solve the problem, we begin by analyzing the geometry and coordinate relationships within triangle $ ABC $, with the centroid $ G $, and the line through $ G $ parallel to side $ BC $ intersecting sides $ AB $ and $ AC $ at points $ M $ and $ N $, respectively. The goal is to find the area of triangle $ ENG $, where $ E $ is the midpoint of side $ AC $, and the area of triangle $ ABC $ is given as 144.

---

### Step 1: Coordinate System

We assign coordinates to triangle $ ABC $ to simplify the computation:

- Let $ A = (0, 0) $, $ B = (b, 0) $, and $ C = (c, h) $. This sets up a general triangle with base $ AB $ along the x-axis.

- Then, the midpoint $ E $ of $ AC $ is:
  $$
  E = \left( \frac{c}{2}, \frac{h}{2} \right)
  $$

- The centroid $ G $ of triangle $ ABC $ is the average of the coordinates of the vertices:
  $$
  G = \left( \frac{b + c}{3}, \frac{h}{3} \right)
  $$

- Since the line through $ G $ is parallel to $ BC $, and intersects sides $ AB $ and $ AC $, the points $ M $ and $ N $ divide the triangle in a specific ratio along these sides. By the properties of similar triangles and the centroid, we can deduce that this line cuts $ AB $ and $ AC $ at $ \frac{2}{3} $ of the total length from the vertices, i.e.,
  $$
  M = \left( \frac{2b}{3}, 0 \right), \quad N = \left( \frac{2c}{3}, \frac{2h}{3} \right)
  $$

---

### Step 2: Use of Area Formula

Now, we compute the area of triangle $ ENG $ using vector cross product, which is well-suited to this setup.

Let the vectors be:
- $\vec{E} = \left( \frac{c}{2}, \frac{h}{2} \right)$
- $\vec{N} = \left( \frac{2c}{3}, \frac{2h}{3} \right)$
- $\vec{G} = \left( \frac{b + c}{3}, \frac{h}{3} \right)$

Vectors used for area computation:
- $\vec{N} - \vec{E} = \left( \frac{2c}{3} - \frac{c}{2}, \frac{2h}{3} - \frac{h}{2} \right) = \left( \frac{c}{6}, \frac{h}{6} \right)$
- $\vec{G} - \vec{E} = \left( \frac{b}{3} - \frac{c}{2}, \frac{h}{3} - \frac{h}{2} \right) = \left( \frac{2b - 3c}{6}, -\frac{h}{6} \right)$

Now compute the magnitude of the cross product (in 2D, we take the determinant):

$$
\text{Area} = \frac{1}{2} \left| \vec{N} - \vec{E} \right| \times \left| \vec{G} - \vec{E} \right|
$$

But instead of computing determinants directly, we observe a more elegant pattern: all computations show that the area of triangle $ ENG $ is **1/18** of the area of triangle $ ABC $.

---

### Step 3: Final Calculation

Since the area of triangle $ ABC $ is given as 144, the area of triangle $ ENG $ is:

$$
\text{Area of } ENG = \frac{1}{18} \times 144 = \boxed{8}
$$

---

### 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 $72 \cdot (1/3)^2 = \boxed{8}$.