Solution

Given three vertices of a parallelogram $$ABCD$$ are $$A (3, -1, 2), B (1, 2, -4)$$ and $$C (-1, 1, 2)$$ .

Let the coordinates of the fourth vertex be $$D(x, y, z)$$.

We know that the diagonals of a parallelogram bisect each other.

Therefore in parallelogram $$ABCD$$, $$AC$$ and $$BD$$ bisect each other .

$$\displaystyle \therefore $$ Mid-point of $$AC =$$ Mid-point of $$BD$$

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

$$\displaystyle \Rightarrow \left ( 1,0,2 \right )=\left ( \frac{x+1}{2},\frac{y+2}{2},\frac{z-4}{2} \right )$$

$$\displaystyle \Rightarrow \frac{x+1}{2}=1,\frac{y+2}{2}=0$$ and $$\dfrac{z-4}{2}=2$$

$$\displaystyle \Rightarrow x=1, y=-2$$ and $$z=8$$

Thus, the coordinates of the fourth vertex are $$(1, -2, 8)$$