The strategy is more intuitive than anything. Since 3 is a prime number, you know that if any of those individual terms are divisible by 3 (i.e. N, N+1, etc.), than the whole product will be divisible by 3.
The thought process I use is as follows: essentially (without using algebraic expansion) you need to look for 3 consecutive numbers. The reasoning is that if you have N * (N+1) * (N+2), one of the terms must be divisible by 3. None of the cases at first glance fit this criteria.
The next step is the key: you need to determine which answer has 3 terms that would have 3 different remainders when divided by 3. For example, take E: N(N+5)(N-6). The term (N-6) would have the same remainder as the term N, since they have a difference of 6 (also divisible by 3).
Just by looking at the answers mentally, A: N(N+1)(N-4) is the only answer that satisfies this.