Does there always exist one plane onto which any two lines in $\Bbb{R}^3$ can be projected and intersect? Question
Consider the vectorial equations of any two lines in $\Bbb{R}^3$ where the parameter $\lambda$ varies across the entire real line:
$$\pmatrix{x\\y\\z} = {\bf r}(\lambda) = {\bf s} + \lambda{\bf t}$$
$$\pmatrix{x\\y\\z} = {\bf f}(\lambda) = {\bf g} + \lambda{\bf h}$$
Must there exist at least one plane of the form $$\pmatrix{x\\y\\z}={\bf P}(\lambda,\mu) = {\bf a} + \lambda{\bf b} + \mu{\bf c}$$ having a normal unit vector $${\bf n}={{\bf b}\times{\bf c} \over \left\lvert{\bf b}\right\rvert \left\lvert{\bf c}\right\rvert}$$
such that the projections of the lines onto the plane, given by $${\bf r}_\parallel(\lambda)={\bf r}(\lambda) - \bigl( {\bf r}(\lambda)\cdot{\bf n}\bigr){\bf n}$$ $${\bf f}_\parallel(\lambda)={\bf f}(\lambda) - \bigl( {\bf f}(\lambda)\cdot{\bf n}\bigr){\bf n}$$
intersect each other at at least one point? If so, how can I prove this?

My thoughts
I have a feeling that the position vectors ${\bf s}, {\bf g}, {\bf a}$ are irrelevant and that only the direction vectors ${\bf t}, {\bf h}, {\bf b}, {\bf c}$ should matter.
Eventually the matter must amount to supposing some $(\gamma, \eta)\in\Bbb{R}^2$ and then examining the system of parametric equations implied by $$\pmatrix{x\\y\\z}={\bf P}(\gamma, \eta)={\bf r}_\parallel(\gamma)={\bf f}_\parallel(\gamma)$$ perhaps as an augmented matrix.
I feel like if I spent enough time on this, I could brute-force my way through things. However, my practice with vectors in this manner is limited, so there probably is some technique that would greatly facilitate the proof of which I am unaware.
Any tips or ideas?
By the way, I’m asking this question purely out of curiosity. My class has already moved past this unit.
 A: I'm afraid the answer is no, but it's a great question. The key thing to realize is that if you start with parallel lines, the projections onto any plane will either be degenerate (if the plane is perpendicular to the direction vector of the lines) or again parallel. But if you start with skew lines (lines that are neither parallel nor intersecting), you can check that they lie in a unique pair of parallel planes, and so certainly projecting onto one of those, say, you'll get intersecting lines. (Draw a picture!)
EDIT: After a reprimand from the OP (:)), I change my answer to YES. If you start with parallel lines, they determine a unique plane. If you project onto a plane perpendicular to it, but parallel to the original lines, you get a single line. Two lines that coincide entirely are, of course, technically intersecting.
COMMENT: By the way, you should check that if projecting on a plane $\Pi$ yields intersecting lines, for any small perturbation $\Pi'$ of $\Pi$, the same will be true. Of course, it's obvious that it's also always true for any $\Pi'$ parallel to $\Pi$, so only the normal vectors of the planes matter.
A: If the lines are not parallel, the cross product will give a normal to a family of parallel planes.  Choose one and project the original lines onto it.  They will intersect.
