# Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 28

$-6x-3y+3z=-3$

#### 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, $-6x-3y+3z=-3$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.