$12$ Identical balls can be placed into $3$ identical boxes,

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 $$P={N\over 3^{12}}\doteq0.531536\ .$$

Keep in mind that the bags are identical. So if you choose to put $3$ into one bag, then you have to put $9$ into the rest:

3 0 9
3 1 8
3 2 7
3 3 6
3 4 5

If you continued to 3 5 4 then you would have a duplicate, because the bags are identical. So there are $5$ ways to have at least $1$ bag with exactly $3$ balls and $4$ ways to have exactly one bag with exactly $3$ balls.

Since 12 is such a small number, it is easier to just enumerate the placements than to try some counting trick (although they do exist) :

0 0 12   1 1 10   2 2 8    3 3 6    4 4 4
0 1 11   1 2 9    2 3 7    3 4 5
0 2 10   1 3 8    2 4 6
0 3 9    1 4 7    2 5 5
0 4 8    1 5 6
0 5 7    
0 6 6    

So there are $7 + 5 + 4 + 2 + 1 = 19$ total placement options.

Suppose the experiment consists of randomly assigning $12$ balls to $3$ boxes. There are $3^{12}$ ways to do so. We want to find the probability that (at least) one of the boxes contains exactly $3$ balls. Let us number the boxes $1$, $2$ and $3$, and define $E_i$ as the event that box # $i$ contains exactly $3$ balls. We want to find the probability of the event $E_1 \cup E_2 \cup E_3$.

$\Pr(E_1 \cup E_2 \cup E_3) $ $= \Pr(E_1) + \Pr(E_2) + \Pr(E_3) - \Pr(E_1\cap E_2) - \Pr(E_1\cap E_3) - \Pr(E_2\cap E_3) + \Pr(E_1\cap E_2 \cap E_3)$ $= 3\Pr(E_1) - 3\Pr(E_1\cap E_2) + \Pr(E_1\cap E_2 \cap E_3) \ \ \ \ \color{blue}{\text{(By symmetry)}}$

We will now find $\Pr(E_1)$, $\Pr(E_1\cap E_2)$, and $\Pr(E_1\cap E_2 \cap E_3)$.

$\displaystyle \Pr(E_1) = {12\choose 3}\left(\frac{2^{9}}{3^{12}}\right)$

$\displaystyle \Pr(E_1\cap E_2) = {12\choose 3} {9\choose 3}\left(\frac{1^{6}}{3^{12}}\right)$

$\displaystyle \Pr(E_1\cap E_2 \cap E_3) = 0$

Therefore, $\Pr(E_1 \cup E_2 \cup E_3) $ $= 3\Pr(E_1) - 3\Pr(E_1\cap E_2) + \Pr(E_1\cap E_2 \cap E_3) $ $= \displaystyle 3\times {12\choose 3}\left(\frac{2^{9}}{3^{12}}\right) - 3 \times {12\choose 3} {9\choose 3}\left(\frac{1^{6}}{3^{12}}\right)$ $\approx 0.5315$

Here is the code to find the above value. You will also find the computation of the required probability using simulation method in the code: http://www.codeskulptor.org/#user42_g9uaFBlARwEdnL6_1.py