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!

I'd like to add the following to Agustí Roig's splendid answer.
The equation of the general line passing through both $\mathbb{L}_2$ and $\mathbb{L}_3$ is given by $\underline{r} = \underline{a}+ \mu(\mathbb{L}_2 - \mathbb{L}_3),$ where $\underline{a}$ is some point on this line and its direction is $\mathbb{L}_2 - \mathbb{L}_3.$
Now we know that the direction is (1,3,2), so set
$\lambda (1,3,2) = \mathbb{L}_2 - \mathbb{L}_3.$
That is
$$\lambda (1,3,2) = (0,0,-1)+\lambda_2(-1,2,3) – (2,0,1) - \lambda_3(1,1,-2).$$
Hence we obtain $\lambda_2 = -2$, $\lambda_3 = 2$ and $\lambda = -2.$
Putting $\lambda_2 = -2$ in the equation of $\mathbb{L}_2$ gives the point (2,-4,-7), which is equivalent to w=0 in my comment to Agustí's answer. I hope that this adds some value for you.
