Comment on No Transposition in K4 of Kryptos by mark

Just a thought, but maybe you need to complete the square with part of #3
XCA NYO USE
EAN YTH ING
Q?O BKR UOX Etc.

Q? would give 9×11
XCA… would be 9×13

What I’m getting at is 9×12 could be 2 Rubik’s cubes? yikes!