Different box and different ball

There are two possible configurations, assuming that the boxes are all the same. They are: (1) 1, 1, 3 and (2) 1, 2, 2. When taking into the account that the boxes are different, we will have the multiply by 3 for each of the configurations in the end (since $\frac{3!}{2!}=3$ ).

Now for the first configuration, the number of ways is $\left(\begin{array}{c}5\\ 3\end{array}\right)(\text{choose 3 out of 5 balls for the box with 3 balls})\times2(\text{arrangements for the remaining 2 balls})\times3(\text{multiplier})$

For the second configuration, the number of ways is $\left(\begin{array}{c}5\\ 2\end{array}\right)(\text{choose 2 out of 5 balls for one of the boxes with 2 balls})\times\left(\begin{array}{c}3\\ 2\end{array}\right)(\text{arrangements for the remaining 3 balls})\times3(\text{multiplier})$

Hence the answer is 150.