= Solution
We work over an <algebraically closed field> $k$, so the assertion concerns a geometric <projective line>. Let $V=k^4$ and let $W$ be the <vector space> of homogeneous cubic forms on $V$. It has dimension $\binom63=20$, so the parameter space of cubic equations up to nonzero scalar is $\mathbb P(W)=\mathbb P^{19}$. The <Grassmannian> $G=\operatorname{Gr}(2,V)$ parametrizes lines in $\mathbb P^3$. Its standard graph charts are four-dimensional <affine spaces>, so its dimension is four. The <Plücker embedding> realizes it as a closed quadric in $\mathbb P^5$, proving projectivity. The <general linear group> $\operatorname{GL}_4$ acts transitively on two-planes and is an irreducible open subset of the affine space of matrices; its orbit map proves that $G$ is irreducible.
Consider the incidence variety
$$
\mathcal I=\{(\ell,[F])\in G\times\mathbb P(W):F|_\ell=0\}.
$$
Restriction of a cubic to a fixed line is surjective onto the four-dimensional space of binary cubic forms: after choosing coordinates with that line given by $x_2=x_3=0$, extend any binary cubic using $x_0,x_1$ alone. Thus its kernel has dimension sixteen. On each <Grassmannian> chart these restriction maps depend algebraically on the line parameters and have constant rank four. Their kernels form a rank-sixteen <vector bundle>, and $\mathcal I$ is its <projective bundle>, with fibre $\mathbb P^{15}$. In particular, \b[$\mathcal I$ is irreducible of dimension $4+15=19$]. It is closed in the product, since restriction vanishes precisely when four polynomial coefficients vanish.
The projection $\pi:\mathcal I\to\mathbb P(W)$ is a <projective morphism>, hence its image is closed. It remains to prove that this image has dimension nineteen. Equal dimensions alone would not suffice: we must control a fibre locally.
Take
$$
F_0=x_0^2x_2+x_1^2x_3,\qquad \ell_0=\{x_2=x_3=0\}.
$$
In the open <Grassmannian> chart around $\ell_0$, every line is uniquely a graph
$$
x_2=a_0x_0+a_1x_1,\qquad x_3=b_0x_0+b_1x_1.
$$
Restriction of $F_0$ to that line is
$$
a_0x_0^3+a_1x_0^2x_1+b_0x_0x_1^2+b_1x_1^3.
$$
The four <monomials> are linearly independent over every field. Hence the only line in this chart lying on $F_0$ has $a_0=a_1=b_0=b_1=0$. The fibre $\pi^{-1}([F_0])$ therefore has local dimension zero at $(\ell_0,[F_0])$. We do not need $F_0$ to define a smooth surface, nor do we claim its entire fibre is finite.
The <fiber dimension theorem>, in its local dimension inequality, gives
$$
0=\dim_{(\ell_0,[F_0])}\pi^{-1}([F_0])\ge\dim\mathcal I-\dim\pi(\mathcal I).
$$
Thus the closed image has dimension at least nineteen. A proper closed subset of the irreducible $\mathbb P^{19}$ has smaller dimension, so $\pi(\mathcal I)=\mathbb P^{19}$. Every cubic equation consequently has a line in its zero set. In particular, \b[every nonsingular cubic surface contains a line]. The proof in fact establishes existence for arbitrary <cubic surfaces> over an <algebraically closed field>; it does not assert a rational line over an arbitrary nonclosed ground field.
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