Answer
$2x+y+z=5$
Work Step by Step
The normal to the plane is $n=\lt 2,1,1 \gt$
We know that the standard equation of a plane passing through the point $(x_0,y_0,z_0)$ is written as: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Then , we have
$2(x-3)+1(y +2)+1(z-1)=0$
or, $2x-6+y+2+z-1=0$
or, $2x+y+z=5$
