The mid-points of the sides of a triangle are

Solution:

Given the mid-points of the sides of a triangle are $(5,7,11),(0,8,5)$ and $(2,3,-$ 1).

Let he vertices be $A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2}, z_{2}\right)$ and $A\left(x_{3}, y_{3}, z_{3}\right)$ respectively.

Using midpoint formula,

$\Rightarrow\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{2}}{2}\right)=(5,7,11)$

$\Rightarrow \mathrm{x}_{1}=10-\mathrm{x}_{2}, \mathrm{y}_{1}=14-\mathrm{y}_{2}, \mathrm{z}_{1}=22-\mathrm{z}_{2}$

$\Rightarrow\left(\frac{\mathrm{x}_{2}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}\right)=(0,8,5)$

Now consider,

$\Rightarrow \mathrm{x}_{3}=-\mathrm{x}_{2}, \mathrm{y}_{3}=16-\mathrm{y}_{2}, \mathrm{z}_{3}=10-\mathrm{z}_{2}$

$\Rightarrow\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}, \frac{z_{1}+z_{3}}{2}\right)=(0,8,5)$

$\Rightarrow\left(\frac{10-\mathrm{x}_{2}-\mathrm{x}_{2}}{2}, \frac{14-\mathrm{y}_{2}+16-\mathrm{y}_{2}}{2}, \frac{22-\mathrm{z}_{2}+10-\mathrm{z}_{2}}{2}\right)=(2,3,-1)$

$\mathrm{x}_{2}=3, \mathrm{y}_{2}=12, \mathrm{z}_{2}=17$

$x_{1}=10-x_{2}=7, y_{1}=14-y_{2}=2, z_{1}=22-z_{2}=5$

$\therefore \mathrm{x}_{3}=-\mathrm{x}_{2}=-3, \mathrm{y}_{3}=16-\mathrm{y}_{2}=4, \mathrm{z}_{3}=10-\mathrm{z}_{2}=-7$

$\therefore \mathrm{A}(7,2,5), \mathrm{B}(3,12,17), \mathrm{C}(-3,4,-7)$ are the required vertices.

$\Rightarrow \mathrm{x}_{3}=-\mathrm{x}_{2}, \mathrm{y}_{3}=16-\mathrm{y}_{2}, \mathrm{z}_{3}=10-\mathrm{z}_{2}$

$\Rightarrow\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}, \frac{z_{1}+z_{3}}{2}\right)=(0,8,5)$

$\Rightarrow\left(\frac{10-\mathrm{x}_{2}-\mathrm{x}_{2}}{2}, \frac{14-\mathrm{y}_{2}+16-\mathrm{y}_{2}}{2}, \frac{22-\mathrm{z}_{2}+10-\mathrm{z}_{2}}{2}\right)=(2,3,-1)$

$x_{2}=3, y_{2}=12, z_{2}=17$

$\mathrm{x}_{1}=10-\mathrm{x}_{2}=7, \mathrm{y}_{1}=14-\mathrm{y}_{2}=2, \mathrm{z}_{1}=22-\mathrm{z}_{2}=5$

$\therefore \mathrm{x}_{3}=-\mathrm{x}_{2}=-3, \mathrm{y}_{3}=16-\mathrm{y}_{2}=4, \mathrm{z}_{3}=10-\mathrm{z}_{2}=-7$

$\therefore A(7,2,5), B(3,12,17), C(-3,4,-7)$ are the required vertices.