Obtain the equation of the planes parallel to: $\pi_1: x-y-2z=0$ that are tangent to the surface $S:x²+y²+z²+2x+2y-1=0$.
Solution:
Completing the square on the surface equation we have the sphere $S: (x+1)^2+(y+1)²+z²=3$ with center $C=(-1,-1,0)$ and radius $r=\sqrt{3}$. The planes that we are looking for have the equation $\pi: x-y-2z+d=0$, since the planes are tangent to the sphere the distance from the center of the sphere to the plane is equal to the radius of the sphere, so we have: $d(C,\pi)=r$, which equals to $\frac{|-1+1-2\times 0+ d|}{\sqrt{6}}=\sqrt{3}$ giving $d=\sqrt{18}$ or $d=-\sqrt{18}$.
What I don't understand is what it's going on inside this dot product, he seems to be taking the dot product between the center and the normal vector of the plane and adding d, I'm not following the reasoning behind all this.