Chapter 12: Vectors and the Geometry of Space - Section 12.5 - Lines and Planes in Space - Exercises 12.5 - Page 727: 32

$x+6y+z=16$

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; $n=\lt -2,-12,-2 \gt$ For point $(1,2,3)$, we have $(-2)(x-1)+(-12)(y-1)+(-2)(z-3)=0$ or, $-2x+2-12y+12-2z+6=0$ Thus, $x+6y+z=16$