 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.