Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 29

$y+z=3$

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; $n=\lt 0,6,6 \gt$ For point $(-1,2,1)$, we get $(0)(x+1)+(6)(y -2)+(6)(z-1)=0$ or, $6y+6z=18$ Thus, $y+z=3$