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

B

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

* Giả sử \(M\left( {a;b;c} \right) \in \left( \Delta  \right) \Rightarrow \dfrac{{a - 1}}{2} = \dfrac{{b + 1}}{{ - 1}} = \dfrac{c}{1} \Rightarrow \left\{ \begin{array}{l}a + 2b + 1 = 0\,\,\left( 1 \right)\\b + c + 1 = 0\,\,\,\,\,\left( 2 \right)\end{array} \right.\)

\(\begin{array}{l}*\,\,\left\{ \begin{array}{l}\overrightarrow {AM}  = \left( {a - 1;b + 1;c - 2} \right)\\\overrightarrow {BM}  = \left( {a - 2;b + 1;c} \right)\end{array} \right.,\,\,\overrightarrow {AM} .\overrightarrow {BM}  = 0\\ \Leftrightarrow \left( {a - 1} \right)\left( {a - 2} \right) + {\left( {b + 1} \right)^2} + c\left( {c - 2} \right) = 0\,\,\,\left( 3 \right)\end{array}\)

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

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}a =  - 2b - 1\\c =  - b - 1\\\left( { - 2b - 2} \right)\left( { - 2b - 3} \right) + {\left( {b + 1} \right)^2} + \left( { - b - 1} \right)\left( { - b - 3} \right) = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a =  - 2b - 1\\c =  - b - 1\\4{b^2} + 10b + 6 + {b^2} + 2b + 1 + {b^2} + 4b + 3 = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a =  - 2b - 1\\c =  - b - 1\\6{b^2} + 16b + 10 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a =  - 2b - 1\\c =  - b - 1\\\left[ \begin{array}{l}b =  - 1\,\,\left( {tm} \right)\\b =  - \dfrac{5}{3}\,\,\left( {ktm} \right)\end{array} \right.\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b =  - 1\\c = 0\end{array} \right. \Rightarrow M\left( {1; - 1;0} \right)\end{array}\)

Chọn B.