Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics IV / 1 / b

The three lines have the convenient parametric equationsFor $L_1$, substitute $\mathbf{a}=\mathbf{i}$, $\mathbf{b}=\mathbf{j}$ in the plane through a point and a line given by a cross product formula. Its normal vector is $\mathbf{j}+\mathbf{i}\wedge \mathbf{y}=(0,1-y_3,y_2)$. Cyclically permuting the coordinates gives all three planes:Each normal vector is nonzero because $\mathbf{y}$ is not on the corresponding line. Thus these are genuine planes, and each is unique.

Suppose a line $T$ through $\mathbf{y}$ meets each $L_i$. Its intersection with $L_i$ is different from $\mathbf{y}$, so $T$ lies in ${\Pi }_i$. A nonzero direction vector $\mathbf{d}$ for $T$ must therefore satisfyA square matrix admitting a nonzero vector $\mathbf{d}$ with $M\mathbf{d}=0$ has zero determinant. Expanding givesThus the necessary condition isThe condition is not sufficient. Take $\mathbf{y}=(2,1,0)$, which is on none of the three original lines. It satisfies the condition, since $2+1-2=1$. The three plane equations becomeTheir common intersection is the single line $T=\{(s,1,0):s\in \mathbb{R}\}$. Any proposed line through $\mathbf{y}$ meeting all three original lines would have to be this $T$. But $T$ never meets $L_1=\{(s,0,1):s\in \mathbb{R}\}$: their second and third coordinates differ. The zero determinant guarantees a common direction for the three planes; it does not guarantee the required intersections. This is a parallel obstruction to a transversal of three skew lines.