Trong không gian Oxyz, cho hai đường thẳng

* Hướng dẫn giải

Đáp án B

Gọi $M\left(4+3t;1-t;-5-2t\right)$ và $N\left(2+u;-3+3u;u\right)$ suy ra $\overline{MN}=\left(-2+u-3t;-4+3u+t;u+2t+5\right)$

Mặt khác $\left\{\begin{array}{l}\overrightarrow{MN}\perp \overrightarrow{u_{{∆}_1}}\\ \overrightarrow{MN}\perp \overrightarrow{u_{{∆}_2}}\end{array}\right.\iff \left\{\begin{array}{l}3\left(-2+u-3t\right)+4-3u-t-2u-4t-10\\ -2+u-3t-12+9u+3t+u+2t+5=0\end{array}\right.\iff \left\{\begin{array}{l}-2u-14t=12\\ 11u+2t=9\end{array}\right.\iff \left\{\begin{array}{l}u=1\\ t=-1\end{array}\right.$

Suy ra $\overrightarrow{MN}(2;-2;4)$.