Answer
$-6x-3y+3z=-3$
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; $n=\lt -6,-3,3 \gt$
For point $(0,2,1)$, we get
$(-6)(x-0)+(-3)(y -2)+3(z-1)=0$
Thus, $-6x-3y+3z=-3$
