Sum of two triangles in a projective plane modulo a conic

Question

Given a conic $C$ in the complex projective plane, say $C=\{c:=x^2+y^2+z^2=0\}$, and two “triangles” (given as zeros of products of 3 linear forms $\ell=\{ax+by+cz\}$) $\ell_1\ell_2\ell_3$, $\ell'_1\ell'_2\ell'_3$, one can express the cubic form $q:=\ell_1\ell_2\ell_3+\ell'_1\ell'_2\ell'_3$ as $\ell''_1\ell''_2\ell''_3+\ell_0''c$.

Indeed, generically $C$ intersects $Q:=\{q=0\}$ in 6 points, which one can partition into 3 pairs $p_{1j},p_{2j}$, $j=1,2,3$, and take $\ell''_1,\ell''_2,\ell''_3$ defined by each pair of these points: $\{\ell''_j=0\}=p_{1j}p_{2j}$. Then $q$ and $\ell''_1\ell''_2\ell''_3$ differ by an element $\ell''_0c$ of the ideal $(c)\subset\mathbb{C}[x,y,z]$.

My question is how to describe $\ell''_1,\ell''_2,\ell''_3$ — one of (or even all of) the $15=\binom{6}{2}$ choices — geometrically, or algebraically. Ideally, it should in terms of intersections of the triangles with $C$, and some scaling factors.

Answer 1

This is not an answer, but a reformulation of the problem, that may sheds some light on it.

Consider the variety of triangles $$ T := \mathrm{Sym}^3\mathbb{P}^2 \subset \mathbb{P}^9 $$ inside the variety of all plane cubcis, and the plane $$ \Pi_C := C \times \mathbb{P}^2 \subset \mathbb{P}^9 $$ of cubics proportional to $C$. Note that $\Pi_C$ does not intersect $T$. Therefore, the linear projection $$ \pi_C \colon \mathbb{P}^9 \dashrightarrow \mathbb{P}^6 $$ out of $\Pi_C$ induces a regular morphism $$ T \to \mathbb{P}^6. $$ The argument in the question shows that this morphism is generically finite of degree 15. Now, the images of the two given triangles generate a line $L \subset \mathbb{P}^6$, and (assuming that the given triangles are general) its preimage in $T$ is a smooth curve $T_L$. Then the question is to find a point in the curve $T_L$ over a given point in the line $L$.