If the mid points of the sides of a triangle AB, BC and CA are D(1, 2, -3), E(3, 0, 1) and F(-1,1,-4), then find the centroid of the triangle.

Answer

To find the centroid of the triangle given the midpoints of its sides, we can follow these steps:

### Step 1: Understand the Midpoints
We are given the midpoints of the sides of triangle ABC:
- D(1, 2, -3) is the midpoint of side AB.
- E(3, 0, 1) is the midpoint of side BC.
- F(-1, 1, -4) is the midpoint of side CA.

### Step 2: Set Up the Midpoint Equations
Using the midpoint formula, we can establish equations for the coordinates of points A, B, and C based on the midpoints.

1. For midpoint D:
\[ D = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) = (1, 2, -3) \]
This gives us:
\[ x_1 + x_2 = 2 \quad (1) \]
\[ y_1 + y_2 = 4 \quad (2) \]
\[ z_1 + z_2 = -6 \quad (3) \]

2. For midpoint E:
\[ E = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2}\right) = (3, 0, 1) \]
This gives us:
\[ x_2 + x_3 = 6 \quad (4) \]
\[ y_2 + y_3 = 0 \quad (5) \]
\[ z_2 + z_3 = 2 \quad (6) \]

3. For midpoint F:
\[ F = \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}, \frac{z_1 + z_3}{2}\right) = (-1, 1, -4) \]
This gives us:
\[ x_1 + x_3 = -2 \quad (7) \]
\[ y_1 + y_3 = 2 \quad (8) \]
\[ z_1 + z_3 = -8 \quad (9) \]

### Step 3: Solve the System of Equations
Now we have a system of equations to solve for \(x_1, x_2, x_3\), \(y_1, y_2, y_3\), and \(z_1, z_2, z_3\).

#### Solving for x-coordinates:
From equations (1), (4), and (7):
1. \(x_1 + x_2 = 2\) (1)
2. \(x_2 + x_3 = 6\) (4)
3. \(x_1 + x_3 = -2\) (7)

From (1) we can express \(x_2\):
\[ x_2 = 2 - x_1 \quad (10) \]

Substituting (10) into (4):
\[ (2 - x_1) + x_3 = 6 \implies x_3 = 4 + x_1 \quad (11) \]

Substituting (11) into (7):
\[ x_1 + (4 + x_1) = -2 \implies 2x_1 + 4 = -2 \implies 2x_1 = -6 \implies x_1 = -3 \]

Now substituting \(x_1\) back into (10) and (11):
\[ x_2 = 2 - (-3) = 5 \]
\[ x_3 = 4 + (-3) = 1 \]

#### Solving for y-coordinates:
From equations (2), (5), and (8):
1. \(y_1 + y_2 = 4\) (2)
2. \(y_2 + y_3 = 0\) (5)
3. \(y_1 + y_3 = 2\) (8)

From (2) we can express \(y_2\):
\[ y_2 = 4 - y_1 \quad (12) \]

Substituting (12) into (5):
\[ (4 - y_1) + y_3 = 0 \implies y_3 = y_1 - 4 \quad (13) \]

Substituting (13) into (8):
\[ y_1 + (y_1 - 4) = 2 \implies 2y_1 - 4 = 2 \implies 2y_1 = 6 \implies y_1 = 3 \]

Now substituting \(y_1\) back into (12) and (13):
\[ y_2 = 4 - 3 = 1 \]
\[ y_3 = 3 - 4 = -1 \]

#### Solving for z-coordinates:
From equations (3), (6), and (9):
1. \(z_1 + z_2 = -6\) (3)
2. \(z_2 + z_3 = 2\) (6)
3. \(z_1 + z_3 = -8\) (9)

From (3) we can express \(z_2\):
\[ z_2 = -6 - z_1 \quad (14) \]

Substituting (14) into (6):
\[ (-6 - z_1) + z_3 = 2 \implies z_3 = 8 + z_1 \quad (15) \]

Substituting (15) into (9):
\[ z_1 + (8 + z_1) = -8 \implies 2z_1 + 8 = -8 \implies 2z_1 = -16 \implies z_1 = -8 \]

Now substituting \(z_1\) back into (14) and (15):
\[ z_2 = -6 - (-8) = 2 \]
\[ z_3 = 8 + (-8) = 0 \]

### Step 4: Calculate the Centroid
The coordinates of points A, B, and C are:
- A(-3, 3, -8)
- B(5, 1, 2)
- C(1, -1, 0)

The centroid \(G\) of triangle ABC is given by:
\[ G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right) \]
Calculating each coordinate:
\[ G_x = \frac{-3 + 5 + 1}{3} = \frac{3}{3} = 1 \]
\[ G_y = \frac{3 + 1 - 1}{3} = \frac{3}{3} = 1 \]
\[ G_z = \frac{-8 + 2 + 0}{3} = \frac{-6}{3} = -2 \]

Thus, the centroid \(G\) is:
\[ G(1, 1, -2) \]

### Final Answer
The centroid of the triangle is \(G(1, 1, -2)\).