<s>[INST] Find a plane M such that $l_1\in M (l_1$ is contained in $M$), and $M$ is parallel $l_2$. $l_{1}  :
\left\lbrace
\begin{array}{lcl}
2 x + 3 y + 4 z = 7 && \mbox{  } \\
  4 x +  y - 2 z = -1 && \mbox{  }  \\
\end{array}
\right.$
$l_{2} : \frac{x-2}{20}=\frac{y-2}{10}=\frac{z +1 }{5}$
I have to find a plane M such that $l_1\in M (l_1$ is contained in $M$), and $M$ is parallel $l_2$.
My attempt :
$\begin{pmatrix}
i&j&k\\
2&3&4\\
4&1&-2\\
\end{pmatrix}=-10\hat{i}+20\hat{j}-10\hat{k} \implies a=-10,b=20,c=-10$
$
\left\lbrace
\begin{array}{lcl}
2 x + 3 y + 4 z = 7 && \mbox{  } \\
  4 x +  y - 2 z = -1 && \mbox{  }  \\
\end{array}
\right. \implies x=z-1,y=-2z+3$.
Denote $z=0 \implies x=-1,y=3$
Plan's equation is $ax+by+cz+d=0$ , subtitue $a,b,c,x,y,z \implies d=-70$
$\vec{l_2}=(2,2,-1)+s(20,10,5)$
I dont how to solve   $M$ is parallel $l_2$ , any help is welcome .
 [/INST] You get the direction vector of $l_1: v_1 = (-1,2,-1)$
and the direction vector of $l_2: v_2=(4,2,1)$
You can just cross them to get the normal vector of M. Since both $l_1,l_2$ are $\perp$ to normal vector of M. So
$N_m=v_1\times v_2 = (4,-3,-10)$
Then M is $4x-3y-10z=d$, find d and you are done. (Use any point on $l_1$)
</s>