I'm not sure you're correct (what is k?).
If everything is distinct:
There are $n$ ways to choose the empty cell. There will be exactly one other cell with more than one ball in it; in fact, exactly two balls in it. There are $$\underbrace{ (n-1)}_{\text{choose}\atop\text{urn}}\cdot
\underbrace{( n (n-1)/2)}_{\text{choose the }\atop\text{two balls}}$$ ways to chose and fill this cell. Since exactly one cell is empty, there are $(n-2)!$ ways to distribute the remaining $n-2$ balls (the cells are being thought of as distinct, and each of the remaining $n-2$ cells gets exactly one of the remaining $n-2$ balls; so the number of ways of distributing the remaining $n-2$ balls is just the number of ways to arrange $n-2$ objects).
So the probability is
$${n\cdot( (n-1) n(n-1)/2)(n-2)!\over n^n}={(n-1)(n-1)!\over2\, n^{n-2}}.$$
Take the simplest model: One after the other of the balls is thrown at random into one of the boxes $a$, $b$, $c$. There are $3^{12}$ different possible histories for that, all of them equiprobable. The number of histories leading to a particular final content $(r,s,t)$ of the three boxes is the coefficient of the term $a^r b^s c^t$ in the expansion of $(a+b+c)^{12}$, i.e., is given by ${12!\over r!\>s!\>t!}$.
There is the question whether "one of the boxes" means "at least one of the boxes", or "exactly one of the boxes". Since one sentence later they talk about "exactly three balls" my working hypothesis is that "at least one of the boxes" is meant.
For the probability in question we have to consider the contents
$$(3,9,0), (3,8,1), (3,7,2), (3,5,4)$$
each of them in six orders, and $(6,3,3)$ in three orders. The total number $N$ of "admissible" histories is therefore given by
$$N=6{12!\over3!}\left({1\over9!}+{1\over8!}+{1\over 7!\>2!}+{1\over 5!\>4!}\right)+3{12!\over 6!\>3!\>3!}=282\,480\ ,$$
and the required probability $P$ is $$P={N\over 3^{12}}\doteq0.531536\ .$$
Best Answer
If no box is empty, one box must contain $2$ balls, and each of the other boxes must contain $1$ ball. There are $n-1$ ways to choose the box that gets $2$ balls, and after that everything is completely determined, so there are just $n-1$ distinguishable arrangements of the balls that have no empty box.
The remainder of the problem is counting all of the possible arrangements. That is a basic stars and bars problem, and the linked Wikipedia article has a decent explanation. Here you want Theorem two; the $k$ of the theorem is your $n-1$, and the $n$ of the theorem is your $n$, so the number of arrangements is
$$\binom{n+(n-1)-1}{n}=\binom{2n-2}n\;.$$