Answer
$3x-2y-z=-3$
Work Step by Step
The plane passing through the point $P(x_0,y_0,z_0)$ and normal to vector $n=ai+bj+ck$ is represented by a component equation such as:
$a(x-x_0)+b(y-y_0)+c(z-z_0)=0~~~~~$ (1)
Since, we have $P(0,2,-1)$ and $n=3i-2j-k$
Thus, from the equation (1), we have
$3(x-0)+(-2)(y-2)+(-1)(z-(-1))=0$
or, $3x-2y+4-z-1=0$
Hence, $3x-2y-z=-3$
