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Use coordinates in which $l=\{x_2=x_3=0\}$ and write the defining cubic aswith 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 varietiesIndeed, wherever $Q_2\ne 0$, the surface equation gives $[x_2:x_3]=[-Q_3:Q_2]$, which remains regular on $l$; the analogous formula works wherever $Q_3\ne 0$.

For the plane with $x_2=sw$, $x_3=tw$, its plane section is $l$ together with the residual plane conicThese 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 $(s,t)$ areThus 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 $z_0^{\mathsf{T}}{M}^{\prime }(\tau )z_0\ne 0$. Since the adjugate matrix of a rank-two symmetric matrix is a nonzero scalar multiple of $z_0{z}_0^{\mathsf{T}}$, the determinant derivative is nonzero. If the rank were at most one, the projective kernel would contain a line, on which the quadratic $z^{\mathsf{T}}{M}^{\prime }z$ 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 $l_i,l_i^{\prime }$. 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 $l\cup l_i\cup l_i^{\prime }$, proving the required coplanarity and giving five distinct pairs.