2001
Problem - 3124
Let $x, y,$ and $z$ be some real numbers such that: $x+2y-z=6$ and $x-y+2z=3$. Find the minimal value of $x^2 + y^2 + z^2$.
Solve these two relations with respect to $z$ leads to $$x=4-z\quad\text{and}\quad y=1+z$$
Therefore,
$$x^2 + y^2+z^2=(4-z)^2 + (1+z)^2+z^2=3z^2-6z+17$$
This is a standard quadratic polynomial whose minimal equals $\boxed{14}$ when $z=1$. Correspondingly, $x=3$ and $y=2$.