= Solution
Use the signature-$(2,2)$ <twistor Hermitian form>
$$
h(Z,W)=\omega^A\overline\rho_A+\pi_{A'}\overline\eta^{A'},\qquad Z=(\omega,\pi),\ W=(\eta,\rho).
$$
A real Minkowski point $x$ is a Hermitian matrix $x^{AA'}$. Every vector of its incidence subspace has the form $(ix\pi,\pi)$. Substitution gives
$$
h((ix\pi,\pi),(ix\rho,\rho))=i\overline\rho\,x\pi-i\overline\rho\,x^\dagger\pi=0.
$$
If the two <null geodesics> meet at $x$, their <null twistors> $X,Y$ both belong to this totally isotropic subspace. In addition to their individual nullness, they must obey
$$
\boxed{h(X,Y)=X^\alpha\overline Y_\alpha=0.}
$$
For independent $X,Y$, this says that every point of the <projective line> $\mathbb P\operatorname{span}(X,Y)$ is null: it lies entirely in <projective null twistor space>. It is the <twistor line> of their meeting point. In the finite affine setting, distinct null directions at the meeting point make their primed spinors independent; proportional primed spinors would give proportional incident twistors and the same null geodesic.
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