My reasoning was to first find the probability that 2 rooms are empty, that is the probability that all students go to the same room : $\displaystyle 3\times\frac{1}{3^{12}}=\frac{1}{3^{11}}$.

Now let's find the probability that one room and only one is empty. Let's find the probability as if it was room 1 that is empty (we'll multiply it by 3 afterwards). This means that there is at least one student in room 2 and one student in room 3, and that the 10 remaining students go in either room.

So we "pick" a student with probability 1/12 to put in room 2, and "pick" another student with probability 1/11 to put in room 3. Then for the 10 remaining students, the probability for them to go in room 2 or 3 is $\displaystyle \left(\frac 23\right)^{10}$.

So for this second probability, I obtain $\displaystyle 3\times \frac{1}{12}\times\frac{1}{11}\times \left(\frac 23\right)^{10}$, which I add to the first probability : $\displaystyle \frac{1}{3^{11}}$.

But this gives 0.0004 and it looks so small...