Let's say I have to find the least distance between origin and the plane $$x-2y-2z = 3$$
I know distance from origin to any $x-y-z$ plane is $\sqrt{x^2 + y^2 + z^2}$
so the constraint will be $$g(x, y, z) = x - 2y - 2z -3$$
however, what will my $f(x,y,z)$ be? Why is it possible to remove the sqrt and make it such that $$ f(x,y,z) = x^2 + y^2 + z^2? $$
On
More generally, we would like to find the point on the hyperplane $\{ \mathrm x \in \mathbb R^n \mid \mathrm a^\top \mathrm x = b \}$, where $\mathrm a \neq {\Bbb 0}_n$, that is closest to the origin. Thus, we have an instance of the least-norm problem
$$\begin{array}{ll} \text{minimize} & \| \mathrm x \|_2^2\\ \text{subject to} & \mathrm a^\top \mathrm x = b\end{array}$$
whose solution is
$$\mathrm x_{\text{LN}} := \color{blue}{\mathrm a \left( \mathrm a^\top \mathrm a \right)^{-1} b}$$
By C-S $$\sqrt{x^2+y^2+z^2}=\frac{1}{3}\sqrt{(1^2+(-2)^2+(-2)^2)(x^2+y^2+z^2)}\geq\frac{1}{3}(x-2y-2z)=1.$$ The equality occurs for $(x,y,z)||(1,-2,-2)$ and $x-2y-2z=3,$ id est, occurs,
which says that we got a minimal value.