Answer
$x-2y+z=6$
Work Step by Step
The standard equation of plane passing through the point $(a,b,c)$ is given as $p(x-a)+q(y-b)+r(z-c)=0$
The equation of normal to the plane; $n=\lt 1,-2,1 \gt$
For point $(1,-2,1)$, we get
$1(x-1)-2(y+2)+1(z-1)=0$
or, $x-1-2y-4+z-1=0$
Thus, $x-2y+z=6$
