Cho \(A\left( {2;0;1} \right),\,\,B\left( {0; - 2;...

C

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

* Giả sử \(M\left( {a;b;c} \right) \in \left( P \right) \Rightarrow 2a - b - c + 4 = 0\).

\(\begin{array}{l}*\,\,\,MA = MB \Leftrightarrow {\left( {a - 2} \right)^2} + {b^2} + {\left( {c - 1} \right)^2} = {a^2} + {\left( {b + 2} \right)^2} + {\left( {c - 3} \right)^2}\\ \Leftrightarrow  - 4a + 4 - 2c + 1 = 4b + 4 - 6c + 9\\ \Leftrightarrow 4a + 4b - 4c + 8 = 0 \Leftrightarrow a + b - c + 2 = 0\,\,\left( 2 \right)\end{array}\)

\(*\,\,MB = 3 \Leftrightarrow {a^2} + {\left( {b + 2} \right)^2} + {\left( {c - 3} \right)^2} = 9\,\,\,\left( 3 \right)\)

* Giải hệ \(\left( 1 \right),\,\,\left( 2 \right),\,\,\left( 3 \right) \Rightarrow \left\{ \begin{array}{l}2a - b - c + 4 = 0\\a + b - c + 2 = 0\\{a^2} + {\left( {b + 2} \right)^2} + {\left( {c - 3} \right)^2} = 9\end{array} \right.\)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}a - 2b + 2 = 0\\a + b - c + 2 = 0\\{a^2} + {\left( {b + 2} \right)^2} + {\left( {c - 3} \right)^2} = 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2b - 2\\2b - 2 + b - c + 2 = 0\\{a^2} + {\left( {b + 2} \right)^2} + {\left( {c - 3} \right)^2} = 9\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 2b - 2\\c = 3b\\{\left( {2b - 2} \right)^2} + {\left( {b + 2} \right)^2} + {\left( {3b - 3} \right)^2} = 9\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a = 2b - 2\\c = 3b\\14{b^2} - 22b + 8 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 0\\b = 1\\c = 3\end{array} \right. \Rightarrow M\left( {0;1;3} \right) \Rightarrow P = a + b + c = 4\end{array}\) 

Chọn C.