It the mid-points of the sides of triangle are (1, 2, -3),
(3, 0, 1) and (2, -2, 5), then the centroid is

Options

A. (0, 0, 0)
B. (1, 2, 3)
C. (2, 0, 1)
D. (2, 3, 0)

Answer

To find the centroid of a 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 a triangle as follows:
- Midpoint D = (1, 2, -3)
- Midpoint E = (3, 0, 1)
- Midpoint F = (2, -2, 5)

### Step 2: Use the Midpoint Formula
The midpoints of the sides of the triangle can be expressed in terms of the vertices of the triangle (let's denote them as A, B, and C). The midpoints can be represented as:
- D (midpoint of AB) = \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)\)
- E (midpoint of BC) = \(\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2}\right)\)
- F (midpoint of AC) = \(\left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}, \frac{z_1 + z_3}{2}\right)\)

### Step 3: Set Up the Equations
From the midpoints, we can set up the following equations:

1. For midpoint D:
- \(\frac{x_1 + x_2}{2} = 1\) → \(x_1 + x_2 = 2\) (Equation 1)
- \(\frac{y_1 + y_2}{2} = 2\) → \(y_1 + y_2 = 4\) (Equation 2)
- \(\frac{z_1 + z_2}{2} = -3\) → \(z_1 + z_2 = -6\) (Equation 3)

2. For midpoint E:
- \(\frac{x_2 + x_3}{2} = 3\) → \(x_2 + x_3 = 6\) (Equation 4)
- \(\frac{y_2 + y_3}{2} = 0\) → \(y_2 + y_3 = 0\) (Equation 5)
- \(\frac{z_2 + z_3}{2} = 1\) → \(z_2 + z_3 = 2\) (Equation 6)

3. For midpoint F:
- \(\frac{x_1 + x_3}{2} = 2\) → \(x_1 + x_3 = 4\) (Equation 7)
- \(\frac{y_1 + y_3}{2} = -2\) → \(y_1 + y_3 = -4\) (Equation 8)
- \(\frac{z_1 + z_3}{2} = 5\) → \(z_1 + z_3 = 10\) (Equation 9)

### Step 4: Solve the System of Equations
Now we will solve these equations step by step.

1. **Solve for x-coordinates:**
- From Equation 1: \(x_1 + x_2 = 2\)
- From Equation 4: \(x_2 + x_3 = 6\)
- From Equation 7: \(x_1 + x_3 = 4\)

Adding Equations 1, 4, and 7:
\[ (x_1 + x_2) + (x_2 + x_3) + (x_1 + x_3) = 2 + 6 + 4 \]
This simplifies to:
\[ 2x_1 + 2x_2 + 2x_3 = 12 \implies x_1 + x_2 + x_3 = 6 \quad (Equation 10) \]

2. **Solve for y-coordinates:**
- From Equation 2: \(y_1 + y_2 = 4\)
- From Equation 5: \(y_2 + y_3 = 0\)
- From Equation 8: \(y_1 + y_3 = -4\)

Adding Equations 2, 5, and 8:
\[ (y_1 + y_2) + (y_2 + y_3) + (y_1 + y_3) = 4 + 0 - 4 \]
This simplifies to:
\[ 2y_1 + 2y_2 + 2y_3 = 0 \implies y_1 + y_2 + y_3 = 0 \quad (Equation 11) \]

3. **Solve for z-coordinates:**
- From Equation 3: \(z_1 + z_2 = -6\)
- From Equation 6: \(z_2 + z_3 = 2\)
- From Equation 9: \(z_1 + z_3 = 10\)

Adding Equations 3, 6, and 9:
\[ (z_1 + z_2) + (z_2 + z_3) + (z_1 + z_3) = -6 + 2 + 10 \]
This simplifies to:
\[ 2z_1 + 2z_2 + 2z_3 = 6 \implies z_1 + z_2 + z_3 = 3 \quad (Equation 12) \]

### Step 5: Find the Centroid
The coordinates of the centroid G of the triangle can be calculated using the formula:
\[ G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3}\right) \]

Substituting the values from Equations 10, 11, and 12:
\[ G = \left(\frac{6}{3}, \frac{0}{3}, \frac{3}{3}\right) = (2, 0, 1) \]

### Final Answer
The centroid of the triangle is \(G = (2, 0, 1)\).

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