Bài toán về điểm và vectơ

Có

$\begin{array}{l}\left\{ \begin{array}{l}\overrightarrow {AB}  = \left( {1 - 1;0 - 2;2 - 5} \right) = \left( {0; - 2; - 3} \right)\\\overrightarrow {AC}  = \left( {4 - 1;7 - 2; - 1 - 5} \right) = \left( {3;5; - 6} \right)\\\overrightarrow {AD}  = \left( {4 - 1;1 - 2;a - 5} \right) = \left( {3; - 1;a - 5} \right)\end{array} \right.\\\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right] = \left( {\left| {\begin{array}{*{20}{c}}{ - 2}&{ - 3}\\5&{ - 6}\end{array}} \right|;\left| {\begin{array}{*{20}{c}}{ - 3}&0\\{ - 6}&3\end{array}} \right|;\left| {\begin{array}{*{20}{c}}0&{ - 2}\\3&5\end{array}} \right|} \right) = \left( {27; - 9;6} \right)\\ \Rightarrow \left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right].\overrightarrow {AD}  = \left( {27; - 9;6} \right).\left( {3; - 1;a - 5} \right) = 60 + 6a\end{array}$

$A,B,C,D$ đồng phẳng khi $\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right].\overrightarrow {AD}  = 0 \Leftrightarrow 60 + 6a = 0 \Leftrightarrow a =  - 10$.

Hai vectơ $\overrightarrow a  = \left( {m;2;3} \right),\overrightarrow b  = \left( {1;n;2} \right)$ cùng phương khi $\frac{m}{1} = \frac{2}{n} = \frac{3}{2} \Leftrightarrow \left\{ \begin{array}{l}m = \frac{3}{2}\\n = \frac{4}{3}\end{array} \right.$

$\Rightarrow 2m + 3n = 7.$