This question looks like difficult logic based but is instead calculation intensive.

There are only three non-zero digits which are multiples of 3: 3, 6, and 9.

Since repetition of digits is not allowed, Using these 3 digits, we can make 3P2 or 3!, i.e. 6 three-digit numbers.

369 396 639 693 936 963

So their sum, S = 369 + 396 + 639 + 693 + 936 + 963 = 3996

We can check that, S is divisible by both 74 and 9 using any division technique.