## Thomas' Calculus 13th Edition

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