abc+ab+bc+ac+a+b+c=164

Rewrite the left-hand side as a factorization: abc + ab + bc + ac + a + b + c = (a+1)(b+1)(c+1) - 1.

So (a+1)(b+1)(c+1) = 165.

If a, b, c are positive integers, then a+1, b+1, c+1 ≥ 2, and we must factor 165 into three integers ≥ 2. Since 165 = 3 · 5 · 11, the only possibility (up to order) is: a+1, b+1, c+1 = 3, 5, 11.

Hence (a, b, c) is a permutation of (2, 4, 10).

Therefore, the positive-integer solutions are all permutations of (2, 4, 10).

If you want nonnegative or all-integer solutions, I can list the additional cases or give a general description.