Câu hỏi 4

\(M\left( {a;b;c} \right) \in \left( {Oxy} \right) \Rightarrow c = 0 \Rightarrow M\left( {a;b;0} \right)\)

\(\begin{array}{l}\Rightarrow 3M{A^2} + 2M{B^2} = 3\left[ {{{\left( {a - 1} \right)}^2} + {b^2} + 9} \right] + 2\left[ {{{\left( {a - 11} \right)}^2} + {{\left( {b + 5} \right)}^2} + {{12}^2}} \right]\\= 3\left( {{a^2} - 2a + {b^2} + 10} \right) + 2\left( {{a^2} - 22a + {b^2} + 10b + 290} \right)\\= 3{a^2} - 6a + 3{b^2} + 30 + 2{a^2} - 44a + 2{b^2} + 20b + 580\\ = 5{a^2} - 50a + 5{b^2} + 20b + 610\\= 5\left( {{a^2} - 10a + {b^2} + 4b + 122} \right)\\= 5\left[ {{{\left( {a - 5} \right)}^2} + {{\left( {b + 2} \right)}^2} + 93} \right] \ge 465\end{array}\)

Dấu bằng xảy ra \( \Leftrightarrow \left\{ \begin{array}{l}a = 5\\b =  - 2\end{array} \right. \Rightarrow P = a + b + c = 5 - 2 + 0 = 3\)

Chọn B.