\(\left(a-1\right)^2+\left(b-2\right)^2+c^2=9\)
Ta có:
\(A=\left|2\left(a-1\right)+\left(b-2\right)-2c+11\right|\le\left|2\left(a-1\right)+\left(b-2\right)-2c\right|+11\)
Và \(\left|2\left(a-1\right)+1.\left(b-2\right)-2c\right|\le\sqrt{\left(4+1+4\right)\left[\left(a-1\right)^2+\left(b-2\right)^2+c^2\right]}=9\)
\(\Rightarrow A\le9+11=20\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=b=3\\c=-2\end{matrix}\right.\) \(\Rightarrow P=...\)