Consider $27$ (pairwise distinct!) lines in $\mathbb{P}^3$ whose intersection graph is that expected¹ of the $27$ lines on a smooth cubic surface. **Question:** Is there a simple necessary and sufficient condition for these $27$ lines to indeed lie on a smooth cubic surface?

For a long time I thought this was always the case, but there is at least one obvious necessary condition²:

(T) Whenever three lines pairwise intersect (i.e., are pairwise coplanar), all three lie on a common plane.

(Because the plane through two mutually intersecting lines on a cubic surface cuts the surface as the union of three distinct lines. There are $45$ such tritangent planes on the cubic surface.)

Is this condition (T) sufficient?

**Further comment and bonus question:** The locally closed subvariety of $\mathrm{Gr}(2,4)^{27}$ (or $\mathrm{Sym}^{27}(\mathrm{Gr}(2,4))$) consisting of configurations of $27$ lines satisfying the incidence conditions (made explicit in note (1) below) is not irreducible (because of the first sentence of note (2) below). What are its irreducible components?

**Notes:**

I.e., we can label the lines as $a_1,\ldots,a_6$, $b_1,\ldots,b_6$ and $c_{12},\ldots,c_{56}$ (the latter indexed by the $15$ unordered pairs in $\{1,\ldots,6\}$) such that the $a_i$ mutually don't intersect, the $b_i$ mutually don't intersect, each $a_i$ intersects each $b_j$ except exactly when $i=j$, the $c_{ij}$ intersect exactly when their index pairs are disjoint, and $c_{ij}$ intersects exactly $a_i,a_j,b_i,b_j$ among the $a_k$ and $b_k$. Equivalently, the intersection graph is the complement of the vertex graph of the Gosset $2_{21}$ polytope. Equivalently, the faithful transitive action of $W(E_6)$ on $27$ elements where the incidence relation is given by the orbit on the unordered pairs such that every line is incident to exactly $10$ others.

Three distinct mutually intersecting lines in $\mathbb{P}^3$ either lie on a common plane or, dually, meet at a common point. (Apologies for pointing out something so obvious, but I'm sure I'm not the only one who might have missed this trivial fact.) If we take the $27$ lines on a smooth cubic surface $X$ and dualize (i.e., replace them by their polars w.r.t. some fixed nondegenerate quadric), we get another configuration of $27$ lines (lying on the projective dual $X^\vee$ to the cubic surface) which have $45$ points of intersection of triples of lines, but in general (unless $X$ had some Eckardt points) no planes in which three lines lie; so this configuration

*does not*lie on a cubic surface (and indeed, $X^\vee$ is not a cubic surface).