Trong không gian với hệ tọa độ Oxyz, cho hai đường thẳng $\left(...

Điều kiện $m\ne 0$
Ta có $\left( {{d}_{1}} \right):\dfrac{x}{3}=\dfrac{-y}{2}=\dfrac{z-2}{m}\Leftrightarrow \dfrac{x}{3}=\dfrac{y}{-2}=\dfrac{z-2}{m}\Rightarrow \left\{ \begin{aligned}
& \!\!\tilde{\mathrm{N}}\!\!\text{ iem A}\left( 0; 0; 2 \right)\in \left( {{d}_{1}} \right) \\
& \text{Vect }\!\!\hat{\mathrm{o}}\!\!\text{ ch ph }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\hat{\mathrm{o}}\!\!\text{ ng }\overrightarrow{u}=\left( 3; -2; m \right) \\
\end{aligned} \right.$
$\left( {{d}_{2}} \right):\dfrac{x+2}{5}=\dfrac{-5-y}{-3}=\dfrac{z}{1}\Leftrightarrow \dfrac{x+2}{5}=\dfrac{y+5}{3}=\dfrac{z}{1}\Rightarrow \left\{ \begin{aligned}
& \!\!\tilde{\mathrm{N}}\!\!\text{ iem B}\left( -2; 5; 0 \right)\in \left( {{d}_{2}} \right) \\
& \text{Vect }\!\!\hat{\mathrm{o}}\!\!\text{ ch ph }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\hat{\mathrm{o}}\!\!\text{ ng }\overrightarrow{{{u}_{2}}}=\left( 5; 3; 1 \right) \\
\end{aligned} \right.$
$\left( {{d}_{1}} \right)$ và $\left( {{d}_{2}} \right)$ cắt nhau $\Leftrightarrow \left\{ \begin{aligned}
& \left[ \overrightarrow{{{u}_{1}}}, \overrightarrow{{{u}_{2}}} \right]\ne \overrightarrow{0} \\
& \left[ \overrightarrow{{{u}_{1}}}, \overrightarrow{{{u}_{2}}} \right].\overrightarrow{AB}=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m\in \mathbb{R} \\
& -19m=19 \\
\end{aligned} \right.\Leftrightarrow m=-1$