1 Elimination Method
Example 1 Given that $x, y, z$ are real numbers, and $x+2y-z=6, x-y+2z=3$. Then, the minimum value of $x^{2}+y^{2}+z^{2}$ is $\qquad$
(2001, Hope Cup Junior High School Mathematics Competition)
Solution: From the given, we can solve for $y=5-x, z=4-x$. Then
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x^{2}+(5-x)^{2}+(4-x)^{2} \\
=3(x-3)^{2}+14 . \\
\because 3(x-3)^{2} \geqslant 0, \quad \therefore x^{2}+y^{2}+z^{2} \geqslant 14 .
\end{array}
$$

Therefore, the minimum value of $x^{2}+y^{2}+z^{2}$ is 14.