Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 872: 36

$x-y-z=4$

The general form of the equation of the plane passing through the point $(a,b,c)$ and having normal vector $\lt l,m,n\gt$is: $l(x-a)+m(y-b)+n(z-c)=0$ Thus, the equation of plane is: $-3(x-6)+3(y+1)+3(z-3)=0$ After simplification, we get $(x-6)-(y+1)-(z-3)=0$ Or, $x-y-z=4$