Answer
$2x+y+z=5$
Work Step by Step
The standard equation of a plane passing through the point $(x_1,y_1,z_1)$ can be defined as:
$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$
The normal vector to the plane is given by: $n=\lt 2,1,1 \gt$
Thus, $2(x-3)+1(y +2)+1(z-1)=0 \implies 2x-6+y+2+z-1=0$
or, $2x+y+z=5$
