Trong không gian Oxyz, cho ...

A

- Hướng dẫn giải

\(\begin{array}{l}\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right] = \left( { - 3; - 6;6} \right)\\{S_{ABC}} = \frac{1}{2}\left| {\left( {\overrightarrow {AB} ;\overrightarrow {AC} } \right)} \right| = \frac{9}{2}\\ \Rightarrow {d_{\left( {M,\left( {ABC} \right)} \right)}} = \frac{{3V}}{S} = \frac{9}{{9/2}} = 2\\\left( {ABC} \right):\left( {x - 0} \right) + 2\left( {y - 1} \right) - 2z = 0\\M \in \left( d \right) \Rightarrow M\left( {2m + 1; - m - 2;2m + 3} \right)\\{d_{\left( {M,\left( {ABC} \right)} \right)}} = 2 \Leftrightarrow \frac{{\left| {\left( {2m + 1} \right) + 2\left( { - m - 3} \right) - 2\left( {2m + 3} \right)} \right|}}{{\sqrt {{1^2} + {2^2} + {{\left( { - 2} \right)}^2}} }} = 2\\ \Leftrightarrow \left| {4m + 11} \right| = 6 \Leftrightarrow \left[ \begin{array}{l}m = \frac{{ - 5}}{4}\\m = \frac{{ - 17}}{4}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}M\left( {\frac{{ - 3}}{2}; - \frac{3}{4};\frac{1}{2}} \right)\\M\left( { - \frac{{15}}{2};\frac{9}{4};\frac{{ - 11}}{2}} \right)\end{array} \right.\end{array}\)