The midpoints of the sides of a triangle are

Solution:

The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4).

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

The mid point of AB is (1,5,-1), therefore

$\frac{x_{2}+x_{1}}{2}=1$

$x_{1}+x_{2}=2 \ldots \ldots \ldots . . e q .1$

$\frac{y_{2}+y_{1}}{2}=5$

$y_{1}+y_{2}=10 \ldots \ldots$ eq. 2

$\frac{\mathrm{Z}_{2}+\mathrm{z}_{1}}{2}=-1$

$z_{1}+z_{2}=-2 \ldots \ldots \ldots$ eq. 3

Mid point of $A C$ is $(2,3,4)$, therefore

$\frac{x_{3}+x_{1}}{2}=2$

$x_{1}+x_{3}=4 \ldots \ldots \ldots \ldots e q .4$

$\frac{y_{3}+y_{1}}{2}=3$

$y_{1}+y_{3}=6 \ldots \ldots e q .5$

$\frac{\mathrm{z}_{3}+\mathrm{z}_{1}}{2}=4$

$z_{1}+z_{3}=8 \ldots \ldots \ldots$ eq. 6

Mid point of BC is (0,4,-2), therefore

$\frac{x_{2}+x_{3}}{2}=0$

$x_{2}+x_{3}=0 \ldots \ldots \ldots \ldots .$ eq. 7

$\frac{y_{2}+y_{3}}{2}=4$

$y_{3}+y_{2}=8 \ldots \ldots .$ eq. 8

$\frac{\mathrm{z}_{2}+\mathrm{z}_{3}}{2}=-2$

$z_{3}+z_{2}=-4 \ldots \ldots . .$ eq. 9

now, adding the equations 1,4 and 7, and divide it by two we get,

$x_{1}+x_{2}+x_{3}=3$

now subtracting 1, 4, 7 individually, we get

$x_{1}=3, x_{2}=-1$ and $x_{3}=1$

now, adding the equations 2,5 and 8, and divide it by two we get,

$y_{1}+y_{2}+y_{3}=12$

now subtracting 1, 4, 7 individually, we get

$y_{1}=4, y_{2}=6$ and $y_{3}=2$

now, adding the equations 3,6 and 9, and divide it by two we get,

$z_{1}+z_{2}+z_{3}=1$

now subtracting 1, 4, 7 individually, we get

$z_{1}=5, z_{2}=-7$ and $z_{3}=3$

therefore, the coordinates are A(3,4,5), B(-1,6,-7) and C(1,2,3).