Giải thích các bước giải:

\(\begin{array}{l}
a)\overrightarrow {{n_1}}  = \left( {1;2; - 1} \right);\overrightarrow {{n_2}}  = \left( {2; - 1;1} \right)\\
 \Rightarrow VTCP\overrightarrow {{u_d}}  = \left[ {\overrightarrow {{n_1}} ;\overrightarrow {{n_2}} } \right] = \left( {1; - 3; - 5} \right)\\
d:\left\{ \begin{array}{l}
x + 2y - z + 3 = 0\\
2x - y + z - 5 = 0
\end{array} \right. \Rightarrow N\left( {0;2;7} \right) \in d\\
 \Rightarrow \overrightarrow {MN}  = \left( { - 1;0;8} \right)\\
\overrightarrow {{n_\alpha }}  = \left[ {\overrightarrow {MN} ,\overrightarrow {{u_d}} } \right] = \left( { - 24; - 3; - 3} \right)\\
PTMP\left( \alpha  \right): - 24\left( {x - 1} \right) - 3\left( {y - 2} \right) - 3\left( {z + 1} \right) = 0\\
 \Leftrightarrow 8x + y + z + 9 = 0\\
b)\overrightarrow {{n_1}}  = \left( {1;1; - 1} \right);\overrightarrow {{n_2}}  = \left( {2; - 1;1} \right)\\
 \Rightarrow VTCP\overrightarrow {{u_d}}  = \left[ {\overrightarrow {{n_1}} ;\overrightarrow {{n_2}} } \right] = \left( {0; - 3; - 3} \right)\\
d:\left\{ \begin{array}{l}
x + y - z + 1 = 0\\
2x - y + z - 1 = 0
\end{array} \right. \Rightarrow N\left( {0;0;1} \right) \in d\\
\overrightarrow {{n_\alpha }}  = \left[ {\overrightarrow {{n_\beta }} ,\overrightarrow {{u_d}} } \right] = \left( { - 3; - 3;3} \right)\\
PTMP\left( \alpha  \right): - 3x - 3y + 3\left( {z - 1} \right) = 0\\
 \Leftrightarrow x + y - z + 1 = 0
\end{array}\)

Câu c em làm tương tự nhé