Let $X=Z(xy-zw)\subset\mathbb{P}^3$. Prove that there is no hypersurface $S\subset\mathbb{P}^3$ such that the intersection of $X$ and $S$ is a line.
My initial idea was to restrict $X$ by an affine map supposing $w=1$, i.e., considering $X=Z(z-xy)\subset\mathbb{A}^3$ and $S=Z(f)$ for some irreducible $f$. If the intersection is a line $Z(l_1, l_2)$ for linear polynomials $l_1,l_2$, then $<f, z-xy>=<l_1,l_2>$, which seems absurd, but I'm not being able to prove it.
Here are some steps, where I leave some details for you to check. Assume $F=0$ intersected $X$ in a line $L$ given by linear equations, $l_1=l_2=0$. Then, in particular, we have $xy-zw=q_1l_1+q_2l_2$, for some linear polynomials $q_i$s. Show that $l_i, q_i$ have no common zeroes (essentially due to smoothness of $X$) and thus the line $M$ given by $q_1=q_2=0$ is disjoint from $L$. Now, $F=0$ intersects $X$ precisely in $L$ implies $F=0$ can not intersect $M$ at all. But, every line intersects every hypersurface, leading to a contradiction. Notice that we showed that there is no such hypersurface meeting $X$ in $L$ even set-theoretically, not ideal theoretically. Another note: the argument comes from intersection theory. Any line in $X$ has self-intersection zero, but any hypersurface intersection is ample and thus has self-intersection positive.