The probability that when  12 balls are distributed among three boxes, the first will contain three balls is,

$\frac{^{12}C_3× 2^9}{3^{12}}$

Since each ball can be put into any one of the three boxes. So, the total number of ways in which 12 balls can be put into three boxes is $3^{12}$.

Out of 12 balls, 3 balls can be chosen in ${^{12}C}_3$ ways. Now, remaining 9 balls can be put in the remaining 2 boxes in 2⁹ ways. So, the total number of ways in which 3 balls are put in the first box and the remaining in other two boxes is ${^{12}C}_3 × 2^9$.

Hence, required probability $=\frac{^{12}C_3 × 2^9}{3^{12}}$