Cho đường thẳng $d:\dfrac{x-2}{-1}=\dfrac{y+1}{-1}=\dfrac{z+1}{1}$...

Đường thẳng $d:\left\{ \begin{aligned}
& -x+2=-y-1 \\
& y+1=-z-1 \\
\end{aligned} \right..$
Gọi $M=d\cap \left( P \right)\Rightarrow M$ là nghiệm hệ: $\left\{ \begin{aligned}
& 2x+y-2z=0 \\
& -x+y+3=0 \\
& y+z+2=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=1 \\
& y=-2 \\
& z=0 \\
\end{aligned} \right..$
Lấy $A\left( 2;-1;-1 \right)\in d$
$\Delta $ có $\vec{u}=\left[ {{{\vec{n}}}_{\left( P \right)}}, \overrightarrow{AM} \right]$ với $\left\{ \begin{aligned}
& {{{\vec{n}}}_{\left( P \right)}}=\left( 2;1;-2 \right) \\
& \overrightarrow{AM}=\left( -1;-1;1 \right) \\
\end{aligned} \right.\Rightarrow \vec{u}=\left( -1;0;-1 \right). $ Vậy $ \Delta :\left\{ \begin{aligned}
& x=1-t \\
& y=-2 \\
& z=-t \\
\end{aligned} \right.$