Points of a linear system on a cubic surface Let $S$ be a generic cubic surface and let $C$ be its intersection with a generic quadric surface.
In the linear system of hyperplane sections of $S$, how many points represent the planes $H$ tangent to $C$ at two points, such that the curve $H \cap S$ is singular?
 A: OK, I'll try to answer your corrected question : you are looking at planes which are bitangent to $C$, and tangent to $S$ somewhere. Now these are two independent properties, so in the dual $(\mathbb{P}^3)^*$ your planes are the intersection points of the dual surface $S^*$ and the curve of bitangent planes, say $\Gamma $.
The degree of $S^*$ is $3.2^2=12$. To compute the degree of $\Gamma $ one counts its intersection with a general plane in $(\mathbb{P}^3)^*$, that is, bitangent planes to $C$ passing  through a general point $p$ of $\mathbb{P}^3$. From $p$ $C$ projects  to a plane sextic $C'$ with 6 nodes, and the bitangent planes through $p$ to bitangent lines to $C'$. Their number is given by the Plücker formula; if I am not mistaken it is 12. So the answer is $12^2=144$. 
A: Zero. The plane cubic $H\cap S$ must have two singular points $p,q$; this implies that the line $\ell:=\langle p,q\rangle$ is contained in $S$. For each point $r$ of $\ell$ the tangent plane $T_r(S)$
cuts $S$ along $\ell$ and a conic $c $; if $\ell\cap c =\{r,r'\} $, we have $T_r(S)=T_{r'}(S)$, and the map $r\mapsto r'$ is an involution of $\ell$. The pairs $(r,r')$ form a curve in $\mathrm{Sym^2}(\ell)$, say $\Gamma _{\ell}$.
Now we want $r,r'$ in a quadric $Q$. This quadric will meet $\ell$ in two points $s,s'$ which are arbitrary; if we choose $Q$ general enough, we will have $(s,s')\notin \Gamma _{\ell}$, and this will hold for all of the 27 lines in $S$.  Thus none of the pairs $(r,r')$ with 
$T_r(S)=T_{r'}(S)$ lies in $C\times C$.
