= Solution
{c}
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 satisfy
$$
M\mathbf d=\mathbf0,\qquad
M=\begin{pmatrix}
0&1-y_3&y_2\\
y_3&0&1-y_1\\
1-y_2&y_1&0
\end{pmatrix}.
$$
A square <matrix> admitting a nonzero <vector> $\mathbf d$ with $M\mathbf d=\mathbf0$ has zero <determinant>. Expanding gives
$$
\begin{aligned}
0=\det M
&=(1-y_1)(1-y_2)(1-y_3)+y_1y_2y_3\\
&=1-y_1-y_2-y_3+y_1y_2+y_2y_3+y_3y_1.
\end{aligned}
$$
Thus the necessary condition is
$$
\boxed{y_1+y_2+y_3-y_1y_2-y_2y_3-y_3y_1=1.}
$$
\b[The 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 become
$$
r_2+r_3=1,\qquad -r_3=0,\qquad 2r_2=2.
$$
Their 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>.
Back to article page
