Answer
$x-2y+z=6$
Work Step by Step
The normal to the plane is $n=\lt 1,-2,1 \gt$
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 for point $(1,-2,1)$, we have
$(1)(x-1)-2(y+2)+1(z-1)=0$
or, $x-1-2y-4+z-1=0$
or, $x-2y+z=6$
