Probability of getting 3 balls in 1st box if 12 balls are distributed randomly among 3 boxes

The first calculation is correct and the second is not. There are indeed ${14 \choose 2} (= {14 \choose 12})$ ways to partition $12$ into an ordered sum of three numbers, but not all of them are equally likely. To take just one example, there is only one way for all $12$ balls to end up in the third box, but there are $12$ ways for $11$ balls to end up in the third box and $1$ ball to end up in the first box. Your second calculation assumes each of the possible partitions is equally likely and that's not so.

The number of balls in the first box is binomially distributed. Each ball is in this box with probability ${1\over3}$ and not in this box with probability ${2\over3}$. The probability that we obtain exactly $3$ balls in the first box then is $${12\choose 3}\cdot\left({1\over3}\right)^3\cdot\left({2\over3}\right)^{12-3}\ ,$$ as you obtained in your first solution.

In your second solution you have counted the number of ways to make a first, second, and third heap with totally $12$ undistinguishable balls, and then you counted how many of these partitions have three balls on the first heap. Looking at the way the experiment is performed in reality there are $3^{12}$ equiprobable allocations of the "secretly numbered" balls to the three boxes, and your first solution counts all allocations having three balls in the first box, whatever their secret number. But the partitions you count as "one" in your second solution are not equiprobable. E.g., all $12$ balls in the third box counts as "one", and so does the partition having four balls in each box. But the latter is much more probable in the real experiment.