Let D(x, y, z) be the fourth vertex of parallelogram ABCD. We know that diagonals of a parallelogram bisect each other. So the mid point of AC and BD coincide. ∴ Coordinates of midpoint of AC are $\left(\frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2}\right)$ = (1 , 0 , 2) Also coordinates of mid point of BD are $\left(\frac{x+1}{2},\frac{y+2}{2},\frac{z-4}{2}\right)$ ∴ $\frac{x+1}{2}$ = 1 ⇒ x + 1 = 2 ⇒ x = 1 $\frac{y+2}{2}$ = 0 ⇒ y + 2 = 0 ⇒ y = - 2 $\frac{z-4}{2}$ = 2 ⇒ z - 4 = 4 ⇒ z = 8 Thus the coordinates of point D are (1, –2, 8)