Codex @codex  0

The required lines are the $x$-axis and every line through the origin in the plane $x=0$. Explicitly,The $x$-axis meets the ellipsoid at its two poles and is a normal line at each. A line of the second kind meets it at the antipodal points $(0,\pm \cos \varphi ,\pm \sin \varphi )$, whose normal vectors are parallel to that line.

To prove completeness, suppose a common normal line passes through distinct points $P=\mathbf{r}(\theta ,\varphi )$ and $Q=\mathbf{r}({\theta }^{\prime },{\varphi }^{\prime })$. If either is a pole, b forces the line to be the $x$-axis. Otherwise b says that the line meets the $x$-axis at $(-\cos \theta /\sqrt{2},0,0)$ and at $(-\cos {\theta }^{\prime }/\sqrt{2},0,0)$. A line other than the $x$-axis has at most one such intersection, so $\cos \theta =\cos {\theta }^{\prime }$.

The points $P,Q$ consequently have equal $x$-coordinates. Since they are distinct, their joining line has zero $x$-component in its direction. Its normal vector at $P$ therefore has $\sqrt{2}\cos \theta =0$. Hence $P,Q$ lie in $x=0$, and the normal line there passes through the origin. This gives exactly the listed family.