= Parallel obstruction to a transversal of three skew lines
For three <skew lines> $L_i$ and a <point> $\mathbf y$ outside them, let $\Pi_i$ be the <plane> through $\mathbf y$ and $L_i$. A <line> through $\mathbf y$ meeting all three $L_i$ must lie in all three $\Pi_i$. Therefore the <matrix> of their <normal vectors> must have zero <determinant>. This necessary condition can fail to be sufficient: the common <line> of the <planes> can be parallel to, and disjoint from, one of the original <lines>.
For example, take $L_1=(s,0,1)$, $L_2=(1,s,0)$, $L_3=(0,1,s)$ and $\mathbf y=(2,1,0)$. The corresponding <planes> have equations $r_2+r_3=1$, $r_3=0$, $r_2=1$. Their common <line> is $(s,1,0)$, which never meets $L_1$. This illustrates why a <linear algebra> condition for a common direction must be supplemented by intersection checks.
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