If points (1,2,3), (0,-4,3), (2,3,5) and (1,-5,-3) are vertices of tetrahedron, then the point where lines joining the mid-points of opposite edges of concurrent is

Options

A. `(1,-1,2)`
B. `(-1,1,2)`
C. (1,1,-2)
D. `(-1,1,-2)`

Explanation

To find the point where the lines joining the midpoints of opposite edges of the tetrahedron are concurrent, we will follow these steps:

### Step 1: Identify the vertices of the tetrahedron
The vertices of the tetrahedron are given as:
- A(1, 2, 3)
- B(0, -4, 3)
- C(2, 3, 5)
- D(1, -5, -3)

### Step 2: Find the midpoints of the edges
We need to find the midpoints of the edges formed by these vertices. The edges we will consider are AB, AC, AD, BC, BD, and CD.

1. **Midpoint of AB**:
\[ M_{AB} = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}, \frac{z_A + z_B}{2} \right) = \left( \frac{1 + 0}{2}, \frac{2 - 4}{2}, \frac{3 + 3}{2} \right) = \left( \frac{1}{2}, -1, 3 \right) \]

2. **Midpoint of AC**:
\[ M_{AC} = \left( \frac{x_A + x_C}{2}, \frac{y_A + y_C}{2}, \frac{z_A + z_C}{2} \right) = \left( \frac{1 + 2}{2}, \frac{2 + 3}{2}, \frac{3 + 5}{2} \right) = \left( \frac{3}{2}, \frac{5}{2}, 4 \right) \]

3. **Midpoint of AD**:
\[ M_{AD} = \left( \frac{x_A + x_D}{2}, \frac{y_A + y_D}{2}, \frac{z_A + z_D}{2} \right) = \left( \frac{1 + 1}{2}, \frac{2 - 5}{2}, \frac{3 - 3}{2} \right) = \left( 1, -\frac{3}{2}, 0 \right) \]

4. **Midpoint of BC**:
\[ M_{BC} = \left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}, \frac{z_B + z_C}{2} \right) = \left( \frac{0 + 2}{2}, \frac{-4 + 3}{2}, \frac{3 + 5}{2} \right) = \left( 1, -\frac{1}{2}, 4 \right) \]

5. **Midpoint of BD**:
\[ M_{BD} = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2}, \frac{z_B + z_D}{2} \right) = \left( \frac{0 + 1}{2}, \frac{-4 - 5}{2}, \frac{3 - 3}{2} \right) = \left( \frac{1}{2}, -\frac{9}{2}, 0 \right) \]

6. **Midpoint of CD**:
\[ M_{CD} = \left( \frac{x_C + x_D}{2}, \frac{y_C + y_D}{2}, \frac{z_C + z_D}{2} \right) = \left( \frac{2 + 1}{2}, \frac{3 - 5}{2}, \frac{5 - 3}{2} \right) = \left( \frac{3}{2}, -1, 1 \right) \]

### Step 3: Find the centroid of the tetrahedron
The centroid (G) of the tetrahedron can be found using the formula:
\[ G = \left( \frac{x_A + x_B + x_C + x_D}{4}, \frac{y_A + y_B + y_C + y_D}{4}, \frac{z_A + z_B + z_C + z_D}{4} \right) \]
Substituting the coordinates:
\[ G = \left( \frac{1 + 0 + 2 + 1}{4}, \frac{2 - 4 + 3 - 5}{4}, \frac{3 + 3 + 5 - 3}{4} \right) \]
Calculating each component:
\[ G = \left( \frac{4}{4}, \frac{-4}{4}, \frac{8}{4} \right) = (1, -1, 2) \]

### Final Answer
The point where the lines joining the midpoints of opposite edges are concurrent is:
\[ \boxed{(1, -1, 2)} \]