1^st ball can go any of the 3 boxes. So total choices for 1^st ball = 3

2^nd ball can also go any of the 3 boxes. So total choices for 2^nd ball = 3
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12^th ball can go any of the 3 boxes. So total choices for 12^th ball = 3

Total choices for all 12 balls = $$3 \times $$$$3 \times $$$$3 \times $$.................12 times = 3¹².

Now question says choose 3 balls from 12 balls. So no of ways = $${}^{12}{C_3}$$ ways.
And then put it in a box. No of ways we can put = $${}^{12}{C_3} \times 1$$ ways.

Now we have 9 balls left and we have to put those 9 balls in the remaining 2 boxes.

Each ball can go to any of the 2 boxes, so for each ball there is 2 choices.

$$\therefore$$ Total ways for 9 balls = 2⁹

$$\therefore$$ Total ways we can put those 12 balls in the boxes = $${}^{12}{C_3} \times 1 \times {2^9}$$

$$\therefore$$ Required probability = $${{{}^{12}{C_3} \times 1 \times {2^9}} \over {{3^{12}}}}$$ = $${{55} \over 3}{\left( {{2 \over 3}} \right)^{11}}$$

So option (C) is correct.