Find 3D line incident on four given 3D lines How can I find the straight line incident on 4 given straight lines, all embedded in 3D space?
I'm usually comfortable with linear algebra; my understanding of projective geometry is limited.
(My ultimate goal is to find a probability distribution for the straight line incident on N given straight lines, where N-1 of them are uncertain/fuzzy. This may be an entirely different question; this question is just to get me started.)
(I will be offline for about 20 hours.)
 A: The other post got too long and the LaTeX starts to lag. 
One can also reason geometrically, directly in $\mathbb{R}^3$ and $\mathbb{RP}^3$. I am going to assume that  $\mathbb{RP}^3$ = $\mathbb{R}^3 \, \cup P_{\infty}$ where $P_{\infty}$ is the plane at infinity. Let $l_1, \, l_2, \, l_3$ and $l_4$ be the four lines such that no two intersect (or are parallel). Take three of them, say $l_1, \, l_2, \, l_3$. Then there is a unique quadratic surface $H$ (aka quadric) in $\mathbb{R}^3$ (and in fact in $\mathbb{RP}^3$) such that all three lines  $l_1, \, l_2, \, l_3$ lie on that quadric $H$. In fact, $H$ is a doubly ruled surfaces, i.e. there is one family of non intersecting lines (call them $l_{\alpha} \, : \, \alpha$) and another family of non intersecting lines (call them $m_{\beta} \, : \, \beta$). Any line $l_{\alpha}$ from the first family intersects any line $m_{\beta}$ from the second family. The three lines $l_1, \, l_2, \, l_3$ are members of the first family $l_{\alpha}$ (they determine it). Think of $H$ as a  one sheeted hyperboloid  (up to projective transformation) or a hyperbolic paraboloid if you wish (wiki these and you will see what I am talking about). 
Now, if you know the three lines in coordinates, you will be able to write explicitly the quadratic equation for $H$.
Assume you have a line $m$ that intersects all three lines $l_1, \, l_2, \, l_3$. Since $m$ intersects $l_1$ and $l_2$, where $l_1 \cap l_2 = \varnothing$, then $m$ intersects $H$ at two different points.  Recall that a line and a qadratic surface can intersect at no more than two points, unless the line lies completely on the surface. However, $m$ intersects also line $l_3$ which is disjoint from the other two lines, so $m$ has a third point of intersection with $H$ so $m$ must lie on $H$. Consequently, $m$ must belong to one of the two families of lines on $H$ and since all lines from $l_{\alpha}$ are disjoint, $m$ is a member of the other family $m_{\beta}$.
Finally, one concludes that if there is a line $m$ that intersects all four lines $l_1, \, l_2, \, l_3$ and $l_4$ it must be a line lying on $H$ so the forth line $l_4$ must have a common point with $H$. The converse is also true, if $l_4$ intersects $H$ at a point $P$, then one can take the unique line $m$ from family $m_{\beta} $ that passes through point $P$. Then $m$ also would intersect the other three lines $l_1, \, l_2, \, l_3$. 
Thus, the solution to your problem boils down to construct the equation of  the quadric $H$, generated by the three lines $l_1, \, l_2, \, l_3$, and to check how many intersection points has $l_4$ with $H$. 
1. If there are two (which is max if all lines are in general position, i.e. if $l_4$ is not a member of $l_{\alpha}$, in which case there would be infinitely many solutions), then there are two lines $m_1$ and $m_2$ that intersect all four lines $l_1, \, l_2, \, l_3$ and $l_4$. 
2. If $l_4$ is tangent to $H$, then there is exactly one  line $m$  that intersects all four lines $l_1, \, l_2, \, l_3$ and $l_4$. 
3. If $l_4$ doesn't intersect $H$ at all, then there is no solution. 
A: I don't think the problem is well defined. If all 4 given lines go through a point O, any other line going through that point is incident on all 4. Similarly, if the 4 lines are co-planar, I can find an infinite number of lines incident on all 4. Now if you have for example 4 parallel lines incident on a plane, and the intersections form a non degenerate quadrilateral, there is no single line that intersects all of them
