Đáp án đúng: A
Phương pháp giải:
Sử dụng công thức: \({V_{ABCD}} = \dfrac{1}{6}\left| {\left[ {\overrightarrow {AB} ;\overrightarrow {AC} } \right].\overrightarrow {AM} } \right|\).Giải chi tiết:Giả sử \(M\left( {a;b;c} \right) \in d\) ta có: \(\dfrac{{a - 1}}{2} = \dfrac{{b + 2}}{{ - 1}} = \dfrac{{c - 3}}{2} \Leftrightarrow \left\{ \begin{array}{l} - a + 1 = 2b + 4\\2b + 4 = - c + 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = - 2b - 3\\c = - 2b - 1\end{array} \right.\).
\( \Rightarrow M\left( { - 2b - 3;b; - 2b - 1} \right)\).
Ta có \(\overrightarrow {AB} = \left( {2;1;2} \right);\,\,\overrightarrow {AC} = \left( { - 2;2;1} \right)\); \(\overrightarrow {AM} = \left( {a;b - 1;c} \right) = \left( { - 2b - 3;b - 1; - 2b - 1} \right)\).
\(\left[ {\overrightarrow {AB} ;\overrightarrow {AC} } \right] = \left( { - 3; - 6;6} \right)\)
\(\begin{array}{l}\left[ {\overrightarrow {AB} ;\overrightarrow {AC} } \right].\overrightarrow {AM} = - 3.\left( { - 2b - 3} \right) - 6\left( {b - 1} \right) + 6\left( { - 2b - 1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 12b + 9\end{array}\)
\( \Rightarrow {V_{ABCD}} = \dfrac{1}{6}\left| {\left[ {\overrightarrow {AB} ;\overrightarrow {AC} } \right].\overrightarrow {AM} } \right| = \dfrac{1}{6}\left| {12b - 9} \right| = \dfrac{1}{2}\left| {4b - 3} \right|\).
Theo bài ra ta có: \(\dfrac{1}{2}\left| {4b - 3} \right| = 3 \Leftrightarrow \left[ \begin{array}{l}b = \dfrac{9}{4}\\b = - \dfrac{3}{4}\end{array} \right. \Rightarrow \left[ \begin{array}{l}M\left( { - \dfrac{{15}}{2};\dfrac{9}{4}; - \dfrac{{11}}{2}} \right)\\M\left( { - \dfrac{3}{2}; - \dfrac{3}{4};\dfrac{1}{2}} \right)\end{array} \right.\) .
Chọn A.