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!

Notation can help us see what's going on.  Let's systematically parametrize the lines: let $\mathbb{L_2}$ start at $\mathbf{a} = (0,0,-1)$ and have a direction vector $\mathbf{u} = (-1,2,3)$ and similarly let $\mathbf{b} = (2,0,1)$, $\mathbf{v} = (1,1,-2)$ determine $\mathbb{L_3}$.  All we need to know about $\mathbb{L_1}$ is its direction vector, $\mathbf{w} = (1,3,2)$.
If there exists a solution then we can start at the origin $\mathbf{a}$ of $\mathbb{L_2}$ and reach the origin $\mathbf{b}$ of $\mathbb{L_3}$ via a path that takes us along $\mathbb{L_2}$, thence in the direction $\mathbf{w}$ of $\mathbb{L_1}$ to a point on $\mathbb{L_3}$, and finally back along $\mathbb{L_3}$ to $\mathbf{b}$.  Let $\lambda$, $\rho$, and $\mu$ be the multiples of the direction vectors $\mathbf{u}$,  $\mathbf{w}$, and  $\mathbf{v}$ that are traveled, respectively.  Whence
$$\mathbf{b - a} = \lambda \mathbf{u} + \rho \mathbf{w} + \mu \mathbf{v}$$
and the solution line $\mathbb{L}$ passes through $\mathbf{a} +  \lambda \mathbf{u}$ in the $\mathbf{w}$ direction.  All we need to do, then, is find $\lambda$.  A simple way is to find a linear form that kills both $\mathbf{w}$ and $\mathbf{v}$ but not $\mathbf{u}$ and apply it to both sides.  The obvious one is to take a dot product with some vector that is mutually perpendicular to $\mathbf{w}$ and $\mathbf{v}$.  All vectors are multiples of the cross product $\mathbf{v} \times \mathbf{w}$, so let's use it.  Thus
$$(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w}) =(\lambda \mathbf{u} + \rho \mathbf{w} + \mu \mathbf{v}) \cdot (\mathbf{v} \times \mathbf{w}) = \lambda \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}).$$
Dividing gives the solution (if it exists!)
$$\lambda = \frac{(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w})}{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})} = -2.$$
A parametrization of the line $\mathbb{L}$ therefore is
$$\mathbf{a} + \mathbf{u} \frac{(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w})}{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})} + t \mathbf{w}$$
for a real parameter $t$.  To check the solution, reverse the roles of the lines $\mathbb{L_2}$ and $\mathbb{L_3}$, thereby finding $\mu$ and the point along $\mathbb{L_3}$.  (We don't have to recompute the triple product in the denominator; only its sign changes.)  The corresponding point (on $\mathbb{L_3}$ now) indeed lies on the solution line (with $t = 2$).
This approach immediately shows a solution does exist and is unique if and only if the triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is nonzero; i.e., the three direction vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are not coplanar.
Of course this all is just another way of reproducing the previous solutions.  Its possible advantages are its geometric clarity and the straightforward formulas it provides for the answer.
