 Solution

Use the signature-$(2,2)$ twistor Hermitian formA real Minkowski point $x$ is a Hermitian matrix $x^{A{A}^{\prime }}$. Every vector of its incidence subspace has the form $(ix\pi ,\pi )$. Substitution givesIf 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 obeyFor independent $X,Y$, this says that every point of the projective line $\mathbb{P}\mathrm{span}(X,Y)$ 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.