There are 4 of each rank.
You can make an A67 flop by combining any of 4 Aces with any of 4 '6's, and then adding any of 4 '7's
that's do-able in 4x4x4 = 64 ways.
There are 52C3 = 22100 ways of picking a 3 card flop.
So p(flop is A67) = 64/22100 = 0.2895%
Same calculation for a KJ2 flop and for a 925 flop
so p(flop is A67 or KJ2 or 925) = 3 x 0.2895% = 0.868%
EDIT: Can also use OddsOracle from ProPokerTools
ProPokerTools Odds Oracle Results (2.25 Professional)
Holdem, Generic syntax
PLAYER_1 *
PLAYER_2 *
600000 trials (randomized)
How often do(es)
the board matches range A67,KJ2,952
0.8613% (5168)
Note that that's using simulation so there's going to be a small difference compared to the answer got from first principles but close enough for sure.
That's giving two players random hands.
If you want to think about giving them blockers, OddsOracle is super-fast at giving you very useable answers.