Three vertices of a parallelogram $ABCD$ are $A(1, 2, 3)$,$B(-1, -2, -1)$ and $C(2, 3, 2)$. Find the fourth vertex $D(x, y, z)$.

(N/A) In a parallelogram,the diagonals bisect each other. Therefore,the mid-point of diagonal $AC$ is the same as the mid-point of diagonal $BD$.
Mid-point of $AC = \left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right) = \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$
Mid-point of $BD = \left(\frac{x-1}{2}, \frac{y-2}{2}, \frac{z-1}{2}\right)$
Equating the mid-points:
$\frac{x-1}{2} = \frac{3}{2}$ $\Rightarrow x-1 = 3$ $\Rightarrow x = 4$
$\frac{y-2}{2} = \frac{5}{2}$ $\Rightarrow y-2 = 5$ $\Rightarrow y = 7$
$\frac{z-1}{2} = \frac{5}{2}$ $\Rightarrow z-1 = 5$ $\Rightarrow z = 6$
Thus,the coordinates of the fourth vertex $D$ are $(4, 7, 6)$.