Let real numbers $ a, b, c$ satisfy $\left\{ \begin{align} & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}\le 25 \\ & 2a+b-2c\ge 12 \\ \end{align} \right.$ . Find the largest value and the smallest value of expression: $$P=a+2b+2c.$$
I solved the problem as below: $$\left\{ \begin{align} & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}\le 25 \\ & 2a+b-2c\ge 12 \\ \end{align} \right.\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-2a-b+2c-13\le 0 (1)$$ $$P=a+2b+2c\Leftrightarrow a+2b+2c-P=0 (2) $$ $(1)$ corresponding sphere with center $I\left( 1;\frac{1}{2};-1 \right)$ and radius $$R=\frac{\sqrt{61}}{2}$$ $(2)$ corresponds to the plane $$\left( \alpha \right):a+2b+2c-P=0$$ The problem of finding conditions of plane and sphere with common intersection. This happens when: $$d\left( I,\left( \alpha \right) \right)=R\Leftrightarrow \frac{\left| -P \right|}{3}\le \frac{\sqrt{61}}{2}\Leftrightarrow -\frac{\sqrt{61}}{2}\le P\le \frac{\sqrt{61}}{2}$$ So $$\min P=-\frac{\sqrt{61}}{2},\max P=\frac{\sqrt{61}}{2}$$ Please comment and give better solution.