Three non-coplanar lines in the 3D-space always have a fourth one that intersect them all? If I have three lines $a,$ $b$ and $c$ in the euclidean 3D space, which are pairwise non-coplanar, is there always a fourth line $x$, that intersects theses three lines?
 A: Label the three lines $\ell_1$, $\ell_2$, and $\ell_3$. They cannot intersect for then they would be co-planar. Since lines $\ell_1$ and $\ell_2$ do not intersect, they have points where they are closest and the line $m$ connecting those two points will be perpendicular to both $\ell_1$ and $\ell_2$. In fact, perpendicular to this line $m$ are two parallel planes: $\mathcal{P}_1$ which contains $\ell_1$ and $\mathcal{P}_2$ which contains $\ell_2$.
Now pick any point $A$ of line $\ell_3$ not in either plane. The point $A$ together with the line $\ell_2$ defines a plane $\mathcal{Q}$ that contains them. This plane, since it intersects plane $\mathcal{P}_2$, must intersect the parallel plane $\mathcal{P}_1$. Moreover, the line $r$ formed by the intersection of planes $\mathcal{Q}$ and $\mathcal{P}_1$ is parallel to line $\ell_2$. Note that lines $r$ and $\ell_1$ are both on $\mathcal{P}_1$ and cannot be parallel. If they were, $\ell_2$ would also be parallel to $\ell_1$, and $\ell_1$ and $\ell_2$ would, therefore, be co-planar. So, lines $r$ and $\ell_1$ must intersect at some point $B$. If you connect points $A$ and $B$ with a line, it will connect point $A$ on $\ell_3$ passes through $\ell_2$ and connect to $B$ on $\ell_1$.
A: I think so, but have not finished the proof.  I would argue that if you pick two of the lines you can find an affine transformation to make the them (x,0,0) and (0,y,1) where x and y parameterize the lines.  The general line through these points is (x-xt,yt,t) where t parameterizes this line.  The third line is (u+u's, v+v's, w+w's).  So to get an intersection, you have three equations in four unknowns: x,y,t,s.  There ought to be a one-parameter family of solutions unless something goes wrong.  You have to show that non-coplanar is enough to make sure nothing goes wrong.
Added:  the equations are $\begin {align} x-xt&=u+u's \\ yt&=v+v's \\ t&=w+w's \end{align}$ 
where $u,u',v,v',w,w'$ are the fixed parameters of the third line.  So you can pick $s$ at will, calculate $t$ from $w, w'$, calculate $y$ from $v, v'$ as long as $t \neq 0$ and calculate $x$ from $u, u'$ as long as $t \neq 1$.  This gives only two values of $s$ that are not acceptable.  So we have the promised one dimensional family of solutions.
A: The answer is yes. Three lines in general position determine a unique hyperboloid of one sheet (since they determine a ruling, or a one parameter family of lines on a hyperbola). But there is a second ruling of the same hyperboloid; every line in this ruling meets the original three lines. So Ross is correct, there is a one parameter family of solutions (in the shape of a hyperboloid).
If I'm ever not up to my eyeballs in homework, I'll make a picture of this and post it.
NOTE: Questions of this sort can be addressed in generality using the Schubert calculus, which is roughly a part of enumerative geometry.
