Finding a line that satisfies three conditions Given lines $\mathbb{L}_1 : \lambda(1,3,2)+(-1,3,1)$, $\mathbb{L}_2 : \lambda(-1,2,3)+(0,0,-1)$ and $\mathbb{L}_3 : \lambda(1,1,-2)+(2,0,1)$, find a line $\mathbb{L}$ such that $\mathbb{L}$ is parallel to $\mathbb{L}_1$, $\mathbb{L}\cap\mathbb{L}_2 \neq \emptyset$ and $\mathbb{L}\cap\mathbb{L}_3 \neq \emptyset$.
Since $\mathbb{L}$ must be parallel to $\mathbb{L}_1$, then $\mathbb{L}:\lambda(1,3,2)+(x,y,z)$ but I can't figure out how to get that (x,y,z) point. I'd like to be given just a slight nod because I'm sure the problem is really easy. Thanks a lot!

$\mathbb L \cap \mathbb L_2 \ne \emptyset$ implies that for some value of $\lambda$, your equation for $\mathbb L$ is on $\mathbb L_2$.  Since substituting $\lambda' = \lambda - c$ doesn't change the line $\mathbb L$, you can take $\lambda = 0$ without loss of generality.  That is, $(x,y,z)$ is on $\mathbb L_2$, or $(x,y,z) = \lambda_2(-1,2,3) + (0,0,-1)$. ...
