Find the equation of the plane through $(1,2,2)$ and parallel to the plane $3x+2y+z=9$

Question

My attempt: The required equation of the plane passing through $(1,2,2)$ is $a(x-1)+b(y-2)+c(z-2)=0$ where $a,b,c$ is the direction ratio's of the line. This equation is parallel to the plane $3x+2y+z=9$ The d.r's of this plane is $a=3,b=2,c=1$ Putting the value of $a,b,c$ in equation we get $$3(x-1)+2(y-2)+1(z-2)=0$$ $$3x-3+2y-4+z-2=0$$ $$3x+2y+z=9$$

Answer 1

The parallel plane must be of the form $3x+2y+z=c$ To find $c$, simply let $(x,y,z)=(1,2,2)$. You get $c = 3(1)+2(2)+(2) = 9$, hence the equation of the plane is $3x+2y+z=9$