The mid point of the sides of a triangle are `(2,3,4),(1,5,-1)` and `(0,4,-2)`. Find the coordinates of the centroid of the triangle.

Answer

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

### Step 1: Identify the midpoints
The midpoints of the sides of the triangle are given as:
- Midpoint D (AC) = (2, 3, 4)
- Midpoint E (BC) = (1, 5, -1)
- Midpoint F (AB) = (0, 4, -2)

### Step 2: Set up the equations for the vertices
Let the vertices of the triangle be A (x1, y1, z1), B (x2, y2, z2), and C (x3, y3, z3).

Using the midpoint formula for three-dimensional coordinates, we can express the midpoints in terms of the vertices:

1. For midpoint D (AC):
\[ \left(\frac{x1 + x3}{2}, \frac{y1 + y3}{2}, \frac{z1 + z3}{2}\right) = (2, 3, 4) \]
This gives us three equations:
\[ \frac{x1 + x3}{2} = 2 \implies x1 + x3 = 4 \quad (1) \]
\[ \frac{y1 + y3}{2} = 3 \implies y1 + y3 = 6 \quad (2) \]
\[ \frac{z1 + z3}{2} = 4 \implies z1 + z3 = 8 \quad (3) \]

2. For midpoint E (BC):
\[ \left(\frac{x2 + x3}{2}, \frac{y2 + y3}{2}, \frac{z2 + z3}{2}\right) = (1, 5, -1) \]
This gives us:
\[ \frac{x2 + x3}{2} = 1 \implies x2 + x3 = 2 \quad (4) \]
\[ \frac{y2 + y3}{2} = 5 \implies y2 + y3 = 10 \quad (5) \]
\[ \frac{z2 + z3}{2} = -1 \implies z2 + z3 = -2 \quad (6) \]

3. For midpoint F (AB):
\[ \left(\frac{x1 + x2}{2}, \frac{y1 + y2}{2}, \frac{z1 + z2}{2}\right) = (0, 4, -2) \]
This gives us:
\[ \frac{x1 + x2}{2} = 0 \implies x1 + x2 = 0 \quad (7) \]
\[ \frac{y1 + y2}{2} = 4 \implies y1 + y2 = 8 \quad (8) \]
\[ \frac{z1 + z2}{2} = -2 \implies z1 + z2 = -4 \quad (9) \]

### Step 3: Solve the equations
Now we have a system of equations to solve for the coordinates of the vertices A, B, and C.

From equation (7):
\[ x2 = -x1 \quad (10) \]

Substituting (10) into (4):
\[ -x1 + x3 = 2 \implies x3 = x1 + 2 \quad (11) \]

Substituting (11) into (1):
\[ x1 + (x1 + 2) = 4 \implies 2x1 + 2 = 4 \implies 2x1 = 2 \implies x1 = 1 \]
Using (10):
\[ x2 = -1 \]
Using (11):
\[ x3 = 1 + 2 = 3 \]

Now for y-coordinates:
From (8):
\[ y2 = 8 - y1 \quad (12) \]
Substituting (12) into (5):
\[ (8 - y1) + y3 = 10 \implies y3 = 2 + y1 \quad (13) \]
Substituting (13) into (2):
\[ y1 + (2 + y1) = 6 \implies 2y1 + 2 = 6 \implies 2y1 = 4 \implies y1 = 2 \]
Using (12):
\[ y2 = 8 - 2 = 6 \]
Using (13):
\[ y3 = 2 + 2 = 4 \]

Now for z-coordinates:
From (9):
\[ z2 = -4 - z1 \quad (14) \]
Substituting (14) into (6):
\[ (-4 - z1) + z3 = -2 \implies z3 = 2 + z1 \quad (15) \]
Substituting (15) into (3):
\[ z1 + (2 + z1) = 8 \implies 2z1 + 2 = 8 \implies 2z1 = 6 \implies z1 = 3 \]
Using (14):
\[ z2 = -4 - 3 = -7 \]
Using (15):
\[ z3 = 2 + 3 = 5 \]

### Step 4: Coordinates of vertices
Now we have the coordinates of the vertices:
- A (1, 2, 3)
- B (-1, 6, -7)
- C (3, 4, 5)

### Step 5: Calculate the centroid
The coordinates of the centroid (G) of a triangle are given by:
\[ G\left(\frac{x1 + x2 + x3}{3}, \frac{y1 + y2 + y3}{3}, \frac{z1 + z2 + z3}{3}\right) \]
Substituting the values:
\[ G\left(\frac{1 + (-1) + 3}{3}, \frac{2 + 6 + 4}{3}, \frac{3 + (-7) + 5}{3}\right) \]
Calculating each coordinate:
\[ G\left(\frac{3}{3}, \frac{12}{3}, \frac{1}{3}\right) = (1, 4, \frac{1}{3}) \]

### Final Answer
The coordinates of the centroid of the triangle are:
\[ (1, 4, \frac{1}{3}) \]