Answer
$x-y+z=0$
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; $n=\lt 3,-3,3 \gt$
For point $(2,1,-1)$, we get
$(3)(x-2)+(-3)(y-1)+(3)(z+1)=0$
or, $3x-3y+3z=0$
Thus, $x-y+z=0$
