If $A$ and $B$ are lines in $\mathbf{CP}^3$ and $P$ is a point in $\mathbf{P}^3$, is there a line incident to $A,B,P$? I want to figure out the following: If $A$ and $B$ are lines in $\mathbf{CP}^3$ and $P$ is a point in $\mathbf{P}^3$ then is it always possible to find a line in $\mathbf{CP}^3$ which meets all of $A,B,P$? I feel that it should be yes, but I fail to translate the problem into a problem in $\mathbf{C}^4$, where I could perhaps use some dimension argument (I mean some linear algebra, the same way as one proves that every two lines in $\mathbf{CP}^2$ intersect). Any ideas?
 A: Consider that the point $P$ does not lie of the first line $l_1$. Then there is a plane passing through this line and the point. This plane then must intersect the second line $l_2$ in at least one point. Now the line through $P$ and the point of intersection is your desired line.
A: Hint. The set of lines in $\mathbf P^3$ which go through your point $P$ is a $\mathbf P^2$. Can you translate the condition «passes through some point of $A$» in terms of that $\mathbf P^2$?
A: For these questions, it is often better to try to think directly, in a geometric way, in $\mathbb P^3$, rather than to translate into a question about $\mathbb C^4$.
In this case, a useful way to think is to consider the span of $A$ and $P$,
which is a plane in $\mathbb P^3$. By definition this is the union of all the lines joining $P$ to a point of $A$, so any line passing through $P$ and meeting $A$ lies in this plane.  In particular, the line you are looking for, if it exists, lies in this plane.
If the line you are looking for exists, then its intersection with $B$ lies both on $B$ and in this plane.  So the next step is to think about how $B$ meets this plane.  Once you've gotten this clear, you should be able to see whether or not you can construct the line you want.
(Note: Mariano's answer gives a more abstract formulation of this approach,
while Rene's answer follows exactly this approach, but is terser.)
