Meaning of $a^2 + b^2 + c^2 = 1$ in a normal form plane equation

Question

As stated above. I know that the $a, b, c$ represent the normal vector of the plane and that you can normalize them so that $a^2 + b^2 + c^2 = 1$. But what is the main reason for doing the normalization?

Answer 1

In the paper, this is done in the context of minimizing the sum of squares of distances to this plane. The distance of a point $(x,y,z)$ from the plane $ax+by+cz+d=0$ is given by the formula $${|ax+by+cz+d|\over\sqrt{a^2+b^2+c^2}}.$$ If you use a unit normal for this plane, for which $a^2+b^2+c^2=1$, then this expression reduces to $|ax+by+cz+d|$, which then allows you to express the square of this distance as the square of the dot product of $[a,b,c,d]$ and $[x,y,z,1]$ as in equation (2) in the paper.