Conic bundle from a line on a smooth cubic surface ($f:S\to {\mathbb{P}}^1,[f^{-1}(t)]=H-L$)

Write a line as $L=\{x_2=x_3=0\}$ and the surface equation as $F=x_2{Q}_2+x_3{Q}_3$. Smoothness says $A=Q_2{|}_L$, $B=Q_3{|}_L$ have no common zero. Projection $[x_2:x_3]$ extends over $L$ by $[-B:A]$. Its fibres are the residual plane conics in the planes through $L$, since substituting $x_2=sw$, $x_3=tw$ gives $F=w(s{Q}_2+t{Q}_3)$. Different residual conics are disjoint everywhere, including their points on $L$, because they are fibres of one morphism of algebraic varieties. The family is flat: over the smooth base curve its local rings are torsion-free over the corresponding discrete valuation ring; smoothness of the total surface supplies the reduced-fibre conclusions.

The matrix of the residual plane conic in coordinates $(u,v,w)$ has coefficient degrees $\left(\begin{array}{ccc}1& 1& 2\\ 1& 1& 2\\ 2& 2& 3\end{array}\right)$ in the plane parameter $(s,t)$. Every term of its determinant has degree five. The reduced singular fibres of a smooth conic bundle lemma rules out an identically zero determinant and makes each zero simple. Over an algebraically closed field, there are therefore exactly five singular fibres, each a pair of distinct coplanar projective lines. No residual conic contains the original line, because $sA+tB\equiv 0$ would make $A,B$ proportional and force a common zero. Pairs from different fibres are disjoint.