Three consecutive vertices of a parallelogram $ABCD$ are $A(6, -2, 4)$,$B(2, 4, -8)$,and $C(-2, 2, 4)$. Find the coordinates of the fourth vertex $D$.

(N/A) Let the coordinates of the fourth vertex $D$ be $(x, y, z)$.
In a parallelogram,the diagonals bisect each other,meaning they share the same midpoint $P$.
Midpoint of diagonal $AC$ is $P\left(\frac{6-2}{2}, \frac{-2+2}{2}, \frac{4+4}{2}\right) = P(2, 0, 4)$.
Midpoint of diagonal $BD$ is $P\left(\frac{x+2}{2}, \frac{y+4}{2}, \frac{z-8}{2}\right)$.
Equating the midpoints:
$\frac{x+2}{2} = 2$ $\Rightarrow x+2 = 4$ $\Rightarrow x = 2$
$\frac{y+4}{2} = 0$ $\Rightarrow y+4 = 0$ $\Rightarrow y = -4$
$\frac{z-8}{2} = 4$ $\Rightarrow z-8 = 8$ $\Rightarrow z = 16$
Thus,the coordinates of the fourth vertex $D$ are $(2, -4, 16)$.