Answer
$7x-5y-4z=6$
Work Step by Step
Since, we are given that the plane is passing through $(1,1,-1),(2,0,2), (0,-2,1)$
Now, $\overrightarrow{p}=\lt 1,-1,3 \gt$ and $\overrightarrow{q}=\lt -1,-3,2 \gt$
The cross product of $\overrightarrow{p}=\lt 1,-1,3 \gt$ and $\overrightarrow{q}=\lt -1,-3,2 \gt$ is equal to $ 7i-5j-4k$
Consider the equation of a plane: $ 7x-5y-4z=D$
For point $(1,1,-1)$; we get
$ 7(1)-5(1)-4(-1)=d \implies d=6 $
Hence, $7x-5y-4z=6$
