What does "scale the coefficients so that $a^2 + b^2 + c^2$ is normalized to 1" mean? I was reading this wikipedia article on plane-sphere intersection plane-sphere intersection and I don't understand what the following means:

Scale the coefficients, if necessary, so that $a^2 + b^2 + c^2$ is
normalized to 1 and d is non-negative.

Can someone please explain what this means (especially "normalizing")? I have some knowledge of basic linear algebra and I know that the norm is the vector perpendicular to a plane. Thanks.
 A: You have the equation $ax+by+cz=d$ of some plane. For any non-zero constant $k$, the equation $akx+bky+ckz=dk$ describes exactly the same plane. This multiplication of every coefficient by $k$ is scaling the coefficients by a factor of $k$. Normalizing something is simply converting it to some sort of standardized value; in this case the something is the sum of the squares of the coefficients of $x,y$, and $z$, and the standardized value is $1$.
The claim, then, is that we can choose $k$ so that $(ak)^2+(bk)^2+(ck)^2=1$, $ak,bk$, and $ck$ being the new scaled coefficients of $x,y$, and $z$. Since
$$(ak)^2+(bk)^2+(ck)^2=k^2(a^2+b^2+c^2)\;,$$
we can do this by setting $$k=\frac1{\sqrt{a^2+b^2+c^2}}\;.$$
If we now let 
$$\begin{align*}
a'&=ak=\frac{a}{\sqrt{a^2+b^2+c^2}},\\
b\,'&=bk=\frac{b}{\sqrt{a^2+b^2+c^2}},\\
c'&=ck=\frac{c}{\sqrt{a^2+b^2+c^2}}\;,\text{ and}\\
d\,'&=dk=\frac{d}{\sqrt{a^2+b^2+c^2}}\;,
\end{align*}$$
we’ll have $(a')^2+(b\,')^2+(c')^2=1$, and the equation $a'x+b\,'y+c'z=d\,'$ describes exactly the same plane as the original $ax+by+cz=d$.
A: Given the equation, with the coefficients $a$, $b$, and $c$ not all zero $$a x + b y + c x = d$$ one can replace $a$ with $$\frac{a}{\sqrt{a^2 + b^2 + c^2}},$$ and similarly for $b$, $c$, and $d$. Now you have a normalized equation,  where $$a^2 + b^2 + c^2 = 1.$$
