= Solution
The meaningful twistor-geometric interpretation of “colinear in $PN$” is that the entire <projective line> through two distinct projective <null twistors> lies in <projective null twistor space>. Mere membership of two points in a <projective line> would impose no condition, because every pair of distinct points in $\mathbb{CP}^3$ spans one.
Indeed, individual nullness gives
$$
h(aX+bY,aX+bY)=a\overline b\,h(X,Y)+b\overline a\,\overline{h(X,Y)}.
$$
This vanishes for every $a,b\in\mathbb C$ precisely when $h(X,Y)=0$. Thus $W=\operatorname{span}(X,Y)$ is a maximal totally isotropic two-plane, a real point of <Minkowski conformal compactification> under the <Lorentzian reality structure on twistor space>. The two corresponding <alpha-planes> meet at this unique $W$. Hence \b[their compact intersection is a real spacetime point], and their real <null geodesics> meet there.
There are two qualifications. The point can be at conformal infinity, so a finite affine meeting is not automatic. If $X,Y$ represent the same projective point instead, their <alpha-planes> coincide rather than having a unique intersection. If “colinear” is read only as spanning a line in $\mathbb{CP}^3$, individual nullness alone does not force a real intersection; the additional orthogonality condition above is indispensable.
For a concrete counterexample to the weaker reading, take $X=(0,0,1,0)$ and $Y=(1,0,0,1)$ in $(\omega^0,\omega^1,\pi_{0'},\pi_{1'})$ coordinates. Each is a <null twistor>, but $h(X,Y)=1$. Their two incidence equations give $x=-i\begin{pmatrix}0&1\\0&0\end{pmatrix}$, which is not Hermitian. Thus their <alpha-planes> meet at a complex point that is not a real Minkowski point.
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