= Solution
{c}
\b[The required <lines> are the $x$-axis and every <line> through the origin in the <plane> $x=0$.] Explicitly,
$$
\boxed{L=\{(t,0,0):t\in\mathbb R\}\quad\text{or}\quad L=\{(0,t\cos\phi,t\sin\phi):t\in\mathbb R\},\quad0\le\phi<\pi.}
$$
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> $(0,\pm\cos\phi,\pm\sin\phi)$, whose <normal vectors> are parallel to that <line>.
To prove completeness, suppose a common <normal line> passes through distinct <points> $P=\mathbf r(\theta,\phi)$ and $Q=\mathbf r(\theta',\phi')$. If either is a pole, <3/b> forces the <line> to be the $x$-axis. Otherwise <3/b> says that the <line> meets the $x$-axis at $(-\cos\theta/\sqrt2,0,0)$ and at $(-\cos\theta'/\sqrt2,0,0)$. A <line> other than the $x$-axis has at most one such intersection, so $\cos\theta=\cos\theta'$.
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 $\sqrt2\cos\theta=0$. Hence $P,Q$ lie in $x=0$, and the <normal line> there passes through the origin. This gives exactly the listed family.
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