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

3 balls for 1^st Box can be selected in ${ }^{12}{\mathrm{C}}_3$ ways, remaining 9 balls can be placed in remaining 2 boxes in 2⁹ ways.
Number of favourable cases ${=}^{12}{\mathrm{C}}_3\times 2^9$

Total number of cases = 3¹²

Required Probability $=\frac{{ }^{12}{\mathrm{C}}_3\times 2^9}{3^{12}}=\frac{{ }^{12}{\mathrm{C}}_3\times 2^9}{9^6}$