- Hướng dẫn giải

$∆$ giao với $d_1: \left\{\begin{array}{l}x=2+t\\ y=1-t\\ z=2-t\end{array}\right.$

là A ( 2 + t; 1 - t; 2 - t )

với $d_2\left\{\begin{array}{l}x=3+t\text{'}\\ y=2+t\text{'}\\ z=5\end{array}\right.$

là B ( 3 + t'; 2 + t'; 5 )

Khi đó AB ( 1 + t' - t; 1 + t' + t; 3 + t )

Ta có:

 $$\begin{gathered} \left\{\begin{array}{l}\overrightarrow{AB}\perp u_1\left(1;-1;-1\right)\\ \overrightarrow{AB}\perp u_2\left(1;1;0\right)\end{array}\right. \\ \iff \left\{\begin{array}{l}1+t\text{'}-t-1-t\text{'}-t-3-t=0\\ 1+t\text{'}-t+1+t\text{'}+t=0\end{array}\right. \\ \iff \left\{\begin{array}{l}t=-1\\ t\text{'}=-1\end{array}\right. \\ \Rightarrow A\left(1;2;3\right),\overrightarrow{AB}\left(1;-1;2\right) \end{gathered}$$

Vậy PT $∆$ là : $d_1:\left\{\begin{array}{l}x=1+t\text{'}\text{'}\\ y=2-t\text{'}\text{'}\\ z=3+2t\text{'}\text{'}\end{array}\right.$

Đáp án cần chọn là A