The three vertices of a parallelogram ABCD are A(-1,3,4), B(2,-1,3) and C(5,1,2). Find the co-ordinates of its 4th vertex D.

Explanation

To find the coordinates of the fourth vertex \( D \) of the parallelogram \( ABCD \) given the vertices \( A(-1, 3, 4) \), \( B(2, -1, 3) \), and \( C(5, 1, 2) \), we can use the property that the diagonals of a parallelogram bisect each other.

### Step-by-Step Solution:

1. **Identify the coordinates of the given points:**
- \( A(-1, 3, 4) \)
- \( B(2, -1, 3) \)
- \( C(5, 1, 2) \)

2. **Find the midpoint \( E \) of diagonal \( AC \):**
The midpoint \( E \) of points \( A \) and \( C \) can be calculated using the midpoint formula:
\[ E = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
For points \( A(-1, 3, 4) \) and \( C(5, 1, 2) \):
\[ E = \left( \frac{-1 + 5}{2}, \frac{3 + 1}{2}, \frac{4 + 2}{2} \right) = \left( \frac{4}{2}, \frac{4}{2}, \frac{6}{2} \right) = (2, 2, 3) \]

3. **Set up the equation for the midpoint \( E \) of diagonal \( BD \):**
Since \( E \) is also the midpoint of \( BD \), we can express this as:
\[ E = \left( \frac{x_B + x_D}{2}, \frac{y_B + y_D}{2}, \frac{z_B + z_D}{2} \right) \]
Substituting the coordinates of \( B(2, -1, 3) \) and \( E(2, 2, 3) \):
\[ (2, 2, 3) = \left( \frac{2 + x_D}{2}, \frac{-1 + y_D}{2}, \frac{3 + z_D}{2} \right) \]

4. **Solve for \( x_D \), \( y_D \), and \( z_D \):**
- For the x-coordinate:
\[ 2 = \frac{2 + x_D}{2} \implies 4 = 2 + x_D \implies x_D = 2 \]
- For the y-coordinate:
\[ 2 = \frac{-1 + y_D}{2} \implies 4 = -1 + y_D \implies y_D = 5 \]
- For the z-coordinate:
\[ 3 = \frac{3 + z_D}{2} \implies 6 = 3 + z_D \implies z_D = 3 \]

5. **Conclusion:**
The coordinates of the fourth vertex \( D \) are:
\[ D(2, 5, 3) \]

### Final Answer:
The coordinates of the fourth vertex \( D \) are \( (2, 5, 3) \).