1. Interpret the experiment

Although the boxes are said to be identical, the given options indicate the standard model where each of the $12$ distinct balls independently goes into one of $3$ boxes with equal probability.

So, total number of outcomes is $3^{12}.$

We need the probability that at least one box contains exactly $3$ balls.

2. Count favorable outcomes

Let the three boxes be temporarily labeled $B_1,B_2,B_3$ for counting.

We want the number of distributions of $12$ distinct balls into $3$ boxes such that one box has exactly $3$ balls.

Since if one box has $3$ balls, the remaining $9$ balls go into the other two boxes.

Step 2.1: Choose the box containing exactly 3 balls

There are $3$ ways.

Step 2.2: Choose which 3 balls go into that box

There are $\binom{12}{3} = 220$ ways.

Step 2.3: Distribute the remaining 9 balls into the other two boxes

Each of the remaining $9$ balls can go to either of the other two boxes, so there are $2^9$ ways.

Thus, total count from this approach is $3\cdot \binom{12}{3}\cdot 2^9.$

But this overcounts the cases where two boxes each contain exactly $3$ balls.

3. Subtract overcounted cases

If two boxes contain exactly $3$ balls each, then the third box contains $6$ balls.

Count such arrangements:

• Choose the box with $6$ balls: $3$ ways.
• Choose the $6$ balls for that box: $\binom{12}{6}$ ways.
• Of the remaining $6$ balls, choose $3$ for one of the two boxes: $\binom{6}{3}$ ways.

So the number of such distributions is $3\binom{12}{6}\binom{6}{3}.$

But this is equivalent to $\frac{3\cdot 12!}{6!3!3!}.$

A simpler inclusion-exclusion setup is better.

4. Use inclusion-exclusion properly

Let $E_i$ be the event that box $i$ contains exactly $3$ balls.

We want $P(E_1\cup E_2\cup E_3).$

So, $|E_1\cup E_2\cup E_3| = \sum |E_i| - \sum |E_i\cap E_j|.$

(Intersection of all three is impossible since $3+3+3\neq 12$.)

Step 4.1: Compute $|E_1|$

Choose 3 balls for box 1, and each of the remaining 9 balls can go to box 2 or 3: $|E_1|=\binom{12}{3}2^9.$

Thus, $\sum |E_i| = 3\binom{12}{3}2^9.$

Step 4.2: Compute $|E_1\cap E_2|$

Choose 3 balls for box 1: $\binom{12}{3}$ Choose 3 of remaining 9 for box 2: $\binom{9}{3}$ The remaining 6 go to box 3.

Hence, $|E_1\cap E_2| = \binom{12}{3}\binom{9}{3}.$

Therefore, $\sum |E_i\cap E_j| = 3\binom{12}{3}\binom{9}{3}.$

So favorable outcomes are $3\binom{12}{3}2^9 - 3\binom{12}{3}\binom{9}{3}.$

5. Compute the probability

$P=\frac{3\binom{12}{3}2^9 - 3\binom{12}{3}\binom{9}{3}}{3^{12}}.$

Factor: $P=\frac{3\binom{12}{3}\left(2^9-\binom{9}{3}\right)}{3^{12}}.$

Now, $\binom{12}{3}=220, \qquad \binom{9}{3}=84, \qquad 2^9=512.$

So, $2^9-\binom{9}{3}=512-84=428.$

Thus, $P=\frac{3\cdot 220\cdot 428}{3^{12}}.$

This does not match any option, so the intended interpretation must be different.

6. Correct interpretation from the options

The options match the probability that a specified box contains exactly 3 balls.

For one fixed box:

• choose which 3 of the 12 balls go into it: $\binom{12}{3}$ ways,
• each of the remaining 9 balls must go into one of the other 2 boxes: $2^9$ ways.

Hence

Simplify:

Now, $\frac{2^9}{3^{12}}=\frac{1}{27}\left(\frac{2}{3}\right)^9\frac{1}{3^3}$ but better simplify directly toward the options:

Rewrite: