#### Answer

$3x-2y-z=-3$

#### Work Step by Step

When the plane passing through the point $P(a,b,c)$ and normal to vector $n=pi+qj+rk$ then we have a component equation as follows:
$a(x-a)+b(y-b)+c(z-c)=0$ ...(x)
Here $P(0,2,-1)$ and $n=3i-2j-k$
From Equation (x), we have
$3(x-0)+(-2)(y-2)+(-1)(z-(-1))=0$
$\implies 3x-2y+4-z-1=0$
Thus, $3x-2y-z=-3$