= Solution
Use coordinates in which $l=\{x_2=x_3=0\}$ and write the defining cubic as
$$
F=x_2Q_2+x_3Q_3
$$
with <homogeneous polynomials> $Q_2,Q_3$ of degree two. Their restrictions $A=Q_2|_l$, $B=Q_3|_l$ have no common zero: at such a zero all first partial derivatives of $F$ would vanish. The rational projection $[x_2:x_3]$ therefore extends to a <morphism of algebraic varieties>
$$
f:S\longrightarrow\mathbb P^1,
\qquad f|_l=[-B:A].
$$
Indeed, wherever $Q_2\ne0$, the surface equation gives $[x_2:x_3]=[-Q_3:Q_2]$, which remains regular on $l$; the analogous formula works wherever $Q_3\ne0$.
For the plane with $x_2=sw$, $x_3=tw$, its <plane section> is $l$ together with the residual <plane conic>
$$
q_{s,t}(u,v,w)=sQ_2(u,v,sw,tw)+tQ_3(u,v,sw,tw)=0.
$$
These residual <plane conics> are exactly the <fibres of a morphism> of $f$, giving the <conic bundle from a line on a smooth cubic surface>. The family itself is smooth: off $l$ it is the graph of projection, and over $l$ the condition $sA+tB=0$ chooses the unique value $[-B:A]$.
Write $q_{s,t}=z^{\mathsf T}M(s,t)z$ with $z=(u,v,w)^{\mathsf T}$ and $M$ symmetric. The degrees of its entries in $(s,t)$ are
$$
\begin{pmatrix}1&1&2\\1&1&2\\2&2&3\end{pmatrix}.
$$
Thus the <conic discriminant> $\Delta(s,t)=\det M(s,t)$ is a <homogeneous polynomial> of degree five.
Smoothness also shows that every zero of $\Delta$ is simple and has <matrix rank> two. To see the first point, at a rank-two singular fibre let $z_0$ span the kernel and use a local base coordinate $\tau$. All fibre-coordinate derivatives vanish at $z_0$, so smoothness of the total family requires $z_0^{\mathsf T}M'(\tau)z_0\ne0$. Since the <adjugate matrix> of a rank-two symmetric matrix is a nonzero scalar multiple of $z_0z_0^{\mathsf T}$, the determinant derivative is nonzero. If the rank were at most one, the projective kernel would contain a line, on which the quadratic $z^{\mathsf T}M'z$ has a zero over the <algebraically closed field>; that point would be singular in the total family. This proves the <reduced singular fibres of a smooth conic bundle> assertion and also rules out $\Delta$ vanishing identically.
The <discriminant quintic of a cubic surface conic bundle> consequently has five distinct zeros in $\mathbb P^1$. At each, a rank-two <plane conic> splits into two distinct <projective lines> $l_i,l_i'$. Neither equals $l$: that would require $sA+tB$ to vanish identically, making $A,B$ proportional, contrary to their having no common zero. The original <plane section> is therefore $l\cup l_i\cup l_i'$, proving the required coplanarity and giving \b[five distinct pairs].
Back to article page
