Hartshorne Exercise V.1.4.
(a) If a surface $X$ of degree $d$ in $\PP{k}{3}$ contains a straight line $C=\PP{k}{1}$, show that $C^2=2-d$.
(b) Assume $\op{char}k=0$, and show for every $d\ge 1$, there exists a nonsingular surface $X$ of degree $d$ in $\PP{k}{3}$ containing the line $x=y=0$.
(a) Let $H\subset\PP{k}{3}$ be a hyperplane such that $dH\sim X$ and $H|_X$ is a very ample divisor on $X$.
By the genus formula on surfaces,
$$C^2=-2-C.K_X.$$
Note that $-K_X=(4-d)H|_X$ using adjunction and $K_{\PP{k}{3}}\sim -4H$. In (V, Ex. 1.2) we saw that $\deg C=C.H|_X$, so assuming that by straight line, Hartshorne means $\deg C=1$, we have $(-K_X).C=(4-d)\deg C=4-d$, therefore $C^2=2-d$, as desired.
(b) Consider the hypersurface in $\PP{k}{3}$ defined by the homogeneous equation, $X^d+XW^{d-1}+Z^{d-1}Y+Y^d=0$. It is irreducible by considering it as a polynomial in $X$ and using Eisenstein’s criterion at the prime $(Y,W)$. It is easy to verify that it is smooth using the Jacobian. It contains the line ${X=0,Y=0}$.
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