The mid-point of the sides of a triangle are (1, 5, – 1),

Solution:

Given the mid-point of the sides of a triangle are $(1,5,-1),(0,4,-2)$ and $(2,3,$, 4).

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

Now by using midpoint formula we get

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

$\Rightarrow x_{1}=2-x_{2}, y_{1}=10-y_{2}, z_{1}=-2-z_{2}$

$\Rightarrow\left(\frac{\mathrm{X}_{2}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{2}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}\right)=(0,4,-2)$

$\Rightarrow x_{3}=-x_{2}, y_{3}=8-y_{2}, z_{3}=-4-z_{2}$

$\Rightarrow\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{3}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{3}}{2}, \frac{\mathrm{z}_{1}+\mathrm{z}_{3}}{2}\right)=(2,3,4)$

$\Rightarrow\left(\frac{2-\mathrm{x}_{2}-\mathrm{x}_{2}}{2}, \frac{10-\mathrm{y}_{2}+8-\mathrm{y}_{2}}{2}, \frac{-2-\mathrm{z}_{2}-4-\mathrm{z}_{2}}{2}\right)=(2,3,4)$

Now by comparing the equations to their RHS we get

$\therefore \mathrm{x}_{2}=-1$,

$y_{2}=6$

$z_{2}=-7$

$\therefore \mathrm{x}_{1}=2-\mathrm{x}_{2}=3$,

$y_{1}=10-y_{2}=4$

$z_{1}=-2-z_{2}=5$

$\therefore \mathrm{x}_{3}=-\mathrm{x}_{2}=1$,

$y_{3}=8-y_{2}=2$

$z_{3}=-4-z_{2}=3$

$\therefore \mathrm{A}(-1,6,-7), \mathrm{B}(3,4,5), \mathrm{C}(1,2,3)$ are the required vertices.

Centroid of a triangle is given by the average of the coordinates of its vertices or midpoint of sides.

Centroid is $\left(\frac{1+0+2}{3}, \frac{5+4+3}{3}, \frac{-1-2+4}{3}\right)=(1,4,1 / 3)$