The domination conjecture for pairings of rational curves on K3 surfaces

Let $X$ be a $K3$ surface, and let $M$ and $N$ be rational curves on $X$ meeting transversely at at least two distinct points $r_1$ and $r_2$. Suppose that

$$M^\perp\cap N^\perp\ne 0.$$

Let $({\mathcal X},{\mathcal M},{\mathcal N})$ be a general deformation of $(X,M,N)$: ${\mathcal X}$ is a family of $K3$ surfaces over a quasi-projective curve $\Gamma$, and ${\mathcal M}$ and ${\mathcal N}\subset {\mathcal X}$ are families of rational curves over $\Gamma$ with $({\mathcal X}_0,{\mathcal M}_0,{\mathcal N}_0)=(X,M,N)$. Fix sections $R_1$ and $R_2$ of ${\mathcal X}/\Gamma$ with $R_i\subset {\mathcal M}\cap {\mathcal N}$ and $R_i\cap {\mathcal X}_0=r_i$ for $i=1,2$. Let $h$ be the family version of the map from ${\mathcal M}^\perp\cap {\mathcal N}^\perp$ to ${\mathbb C}^*$, with

$${\mathcal M}^\perp\cap {\mathcal N}^\perp=\{\gamma\in\operatorname{Pic}({\mathcal X}/\Gamma):\gamma\cdot{\mathcal M}=\gamma\cdot{\mathcal N}=0\}.$$

Domination conjecture. The variety ${\mathbb C}^*$ is dominated by one of the components of ${\mathcal M}^\perp\cap {\mathcal N}^\perp$ under the map $h$.

This conjecture asserts that the family-level pairing map has a component whose image is dominant. The source gives no evidence of a resolution, so its status remains open.