Condition for every line through a line segment $S$ to be perpendicular to $S$?

Question

In 3 dimensional space, given a line segment $S$ which has a line $l(0)$ passing through one endpoint at right angles and another, $l(1)$, passing through the other endpoint at right angles, such that $l(0)$ is not parallel to $l(1)$, prove that the lines passing through each point $l(x)$, $0<x<1$, will also be perpendicular to $S$ on the assumption that none of the lines through $S$ intersect. It is assumed that the $l(x)$, $0\le x \le 1$, are distributed continuously.

1. I believe I might have a counter example, but it's far from clear.
2. Note that this is true for a ruling line of the one-sheeted hyperboloid (or for those of a hyperbolic paraboloid) and its intersection of its dual ruling lines.
3. Of course, this condition could be weakened to the two end lines intersecting $S$ at any pair of congruent angles as long as their concave sides are facing the same directions.
4. If these conditions fail, are there stronger conditions which will make it work?

Answer 1

A counterexample: suppose that the line segment is on the $x$-axis, starting at $(0,0,0)$ and ending at $(\pi,0,0)$. Let $\alpha(t) = \frac\pi4 + \frac12|t - \frac \pi2|$. Through the point $(t,0,0)$, draw the line with unit direction vector $(\sin \alpha(t), \cos t \cos \alpha(t), -\sin t \cos \alpha(t))$.

This vector is orthogonal to the vector $(0, \sin t, \cos t)$, as is the vector $(1,0,0)$, so the entire line we draw lies in the plane with equation $(x,y,z) \cdot (0, \sin t, \cos t) = 0$. As $t$ varies, these planes also vary, repeating with period $\pi$, and they only intersect on the $x$-axis. So as long $t_1 \not\equiv t_2 \pmod \pi$, the line through $(t_1,0,0)$ and the line through $(t_2,0,0)$ cannot intersect; they're skew.

This leaves us with complete freedom to choose the angle that the line makes with the $x$-axis, which we take advantage of: the angle is actually $\alpha(t)$. This is $\frac \pi2$ when $t=0$ or $t=\pi$, but not otherwise.

(I originally wanted to make $\alpha(t) = t + \frac\pi2$, which would be nicer, but the problem we run into is that $\alpha(t)$ should never be $0$. In that case, we end up drawing a line through the line segment, intersecting all the other lines. There's still probably a nicer choice of $\alpha(t)$.)