Cho \(A\left( {1;7;3} \right),\,\,\left( \Delta  \...

C

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

* Giả sử \(M\left( {a;b;c} \right) \in \left( \Delta  \right) \Rightarrow \dfrac{{a - 6}}{{ - 3}} = \dfrac{{b + 1}}{{ - 2}} = \dfrac{{c + 2}}{1} \Leftrightarrow \left\{ \begin{array}{l}a =  - 3c\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\b =  - 2c - 5\,\,\,\left( 2 \right)\end{array} \right.\)

* \(AM = 2\sqrt {30}  \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b - 7} \right)^2} + {\left( {c - 3} \right)^2} = 120\,\,\left( 3 \right)\)

* Giải hệ \(\left( 1 \right),\,\left( 2 \right),\left( 3 \right) \Rightarrow \left\{ \begin{array}{l}a =  - 3c\\b =  - 2c - 5\\{\left( {a - 1} \right)^2} + {\left( {b - 7} \right)^2} + {\left( {c - 3} \right)^2} = 120\end{array} \right.\) 

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}a =  - 3c\\b =  - 2c - 5\\{\left( { - 3c - 1} \right)^2} + {\left( { - 2c - 12} \right)^2} + {\left( {c - 3} \right)^2} = 120\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a =  - 3c\\b =  - 2c - 5\\14{c^2} + 48c + 34 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a =  - 3c\\b =  - 2c - 5\\\left[ \begin{array}{l}c =  - 1\,\,\left( {tm} \right)\\c = \dfrac{{51}}{7}\,\,\left( {ktm} \right)\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b =  - 3\\c =  - 1\end{array} \right. \Rightarrow M\left( {3; - 3; - 1} \right)\end{array}\)

Chọn C.