Let's think through the scenario. First we pick 3 people out of 9. It doesn't matter what order we pick the 3 people in, since they'll all end up in the same group anyway. Since order doesn't matter, we have a combination.

So for this first step, we have: $$\frac{9!}{3!\left(9-3\right)!}=\frac{9!}{3!6!}=\frac{9\cdot8\cdot7}{3\cdot2\cdot1}=3\cdot4\cdot7=84$$

So we have 84 possibilities for how we pick the first group.

Then we need to pick the next group. We pick 3 people out of the remaining 6. Again, it doesn't matter what order we pick them in, so we have a combination:

$$\frac{6!}{3!\left(6-3\right)!}=\frac{6!}{3!3!}=\frac{^{6\cdot5\cdot4}}{3\cdot2\cdot1}=2\cdot5\cdot2=20$$

So for each of the 84 possibilities for how we pick the first group, we have 20 possibilities for how we pick the second group. That's 84*20 = 1680 possibilities in total.

We're left with 3 people. There's only one option at this point, which is to pick all 3 of the remaining people for the last group. So there are 1680 possibilities in total for how we can pick the 3 groups.

HOWEVER, before we pick C, we need to recognize that based on this scenario, we don't care which order the groups themselves are chosen in. This means we need to account for duplicates where the same groups of people are chosen in different orders. (Ex: if the groups are chosen 1.ABC, 2.DEF, 3.GHI, it is the same three groups as when they are chosen 1. GHI, 2.ABC, 3.DEF.) 3 groups can be ordered in 3! ways. So to get rid of any duplicates, we divide our total number of possibilities by 3!.

$$\frac{1680}{3!}=\ 280$$

So the correct answer is A.

Note: this final element of the problem (discounting group ordering) makes this problem particularly challenging. There are other equations by which we can get to the solution faster, but they're so unlikely to come up on a real GMAT exam that they aren't worth memorizing. Using logic and basic combinatorics as we did above is your best bet.