Cho \(A\left( { - 2;1;1} \right),\,\,B\left( { - 3...

B

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

* Giả sử \(M\left( {a;b;c} \right) \in \left( \Delta  \right) \Rightarrow \dfrac{{a + 2}}{1} = \dfrac{{b - 1}}{3} = \dfrac{{c + 5}}{{ - 2}} \Rightarrow \left\{ \begin{array}{l}b = 3a + 7\,\,\,\,\,\,\left( 1 \right)\\c =  - 2a - 9\,\,\left( 2 \right)\end{array} \right.\)

* \(\left\{ \begin{array}{l}\overrightarrow {AM}  = \left( {a + 2;b - 1;c - 1} \right)\\\overrightarrow {AB}  = \left( { - 1; - 2;1} \right)\end{array} \right.;\,\,\overrightarrow n  = \left[ {\overrightarrow {AM} ;\overrightarrow {AB} } \right] = \left( {b + 2c - 3; - a - c - 1; - 2a + b - 5} \right)\)

\({S_{\Delta ABC}} = \dfrac{1}{2}\sqrt {{{\left( {b + 2c - 3} \right)}^2} + {{\left( {a + c + 1} \right)}^2} + {{\left( { - 2a + b - 5} \right)}^2}}  = 3\sqrt 5 \,\,\left( 3 \right)\)

* Giải hệ (1),(2),(3)

\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}b = 3a + 7\\c =  - 2a - 9\\{\left( {b + 2c - 3} \right)^2} + {\left( {a + c + 1} \right)^2} + {\left( { - 2a + b - 5} \right)^2} = 180\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 3a + 7\\c =  - 2a - 9\\{\left( {a + 14} \right)^2} + {\left( {a + 8} \right)^2} + {\left( {a + 2} \right)^2} = 180\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}b = 3a + 7\\c =  - 2a - 9\\3{a^2} + 48a + 84 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}b = 3a + 7\\c =  - 2a - 9\\\left[ \begin{array}{l}a =  - 2\\a =  - 14\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}a =  - 2;b = 1;c =  - 5\,\,\left( {tm} \right)\\a =  - 14;b =  - 35;c = 19\,\,\left( {ktm} \right)\end{array} \right.\end{array}\)

Vậy \(M\left( { - 2;1; - 5} \right)\).

Chọn B.