Let $K$ be any field, and $F=V(f)$ a Zariski closed subset of $K^n$ such that $F\cap L$ is infinite for a given line $L$.
Why does $L\subset F$ hold ?
If $n=2$ for instance, we may assume that the line $L$ is given by $x=0$ for instance, so $f(0,y)$ has infinitely many roots hence it is zero, so $f(x,y)$ vanishes on $L$.
But what for bigger $n$ ? What I understood is that $L\cap F$ is infinite, hence cannot be discrete, by compactness. So it is kind of big... but why as big as $L$ ?
$F\cap L$ is Zariski closed, and the closed subsets of the line are either finite or the line.