Yes. That looks alright.
You counted the ways to partition the objects into groups of distinct sizes $1,2,3,4,5$, and then ways to arrange those groups into the boxes. That is what you wanted to count, and how you could count it.
Also written as $5!\dbinom{15}{5,4,3,2,1}$ using the multinomial coefficient notation.
Let's compare your method with the correct solution.
Solution: As you noted, either three questions are selected from one section with one each from the other two sections or two questions each are selected from two sections and one is selected from the remaining section.
Three questions from one section and one question each from the other two sections: Select the section from which three questions are drawn, select three of the four questions from that section, and select one of the four questions from each of the other two sections. This can be done in
$$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$$
ways.
Two questions each from two sections and one question from the remaining section: Select the two sections from which two questions are drawn, select two of the four questions from each of those sections, and select one of the four questions from the other section. This can be done in
$$\binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1}$$
ways.
Total: Since the two cases are mutually exclusive and exhaustive, there are
$$\binom{3}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\binom{4}{2}\binom{4}{2}\binom{4}{1} = 624$$
ways to select five questions so that at least one is drawn from each of the three sections.
Why your method is wrong?
You are counting each selection in which three questions are drawn from one section and one question is drawn from each of the other sections three times, once for each way you could designate one of the three questions as the question that is drawn from that section. For example, suppose questions $A_1, A_2, A_3, B_1, C_1$ are selected. You count this selection three times.
\begin{array}{c c}
\text{designated questions} & \text{additional questions}\\ \hline
A_1, B_1, C_1 & A_2, A_3\\
A_2, B_1, C_1 & A_1, A_3\\
A_3, B_1, C_1 & A_2, A_3
\end{array}
You are counting each selection in which two questions each are drawn from two sections and one question is drawn from the other section four times, two times for each way you could designate one of the two question from each section from which two questions are drawn as the question that is drawn from that section. For instance, if questions $A_1, A_2, B_1, B_2, C_1$ are drawn, your method counts this selection four times.
\begin{array}{c c}
\text{designated questions} & \text{additional questions}\\ \hline
A_1, B_1, C_1 & A_2, B_2\\
A_1, B_2, C_1 & A_2, B_1\\
A_2, B_1, C_1 & A_1, B_2\\
A_2, B_2, C_1 & A_1, B_1
\end{array}
Notice that
$$\binom{3}{1}\color{red}{\binom{3}{1}}\binom{4}{3}\binom{4}{1}\binom{4}{1} + \binom{3}{2}\color{red}{\binom{2}{1}}\binom{4}{2}\color{red}{\binom{2}{1}}\binom{4}{2}\binom{4}{1} = 2304$$
Best Answer
We could use the Inclusion-Exclusion Principle.
There are $\binom{12}{5}$ ways to select five of the twelve questions on the question paper. From these, we must subtract those selections which do not contain at least one selection from each section.
There are three ways to exclude a section and $\binom{8}{5}$ ways to select five of the remaining eight questions.
Since it is not possible to exclude two of the sections and still select five questions from the remaining four questions, we are done.
$$\binom{12}{5} - \binom{3}{1}\binom{8}{5} = 624$$
Suppose the examinee only had to answer four of the twelve questions, but was still required to answer at least one question from each part.
Observe that the examinee must answer two questions from one part and one from each of the others, which can be done in $$\binom{3}{1}\binom{4}{2}\binom{4}{1}\binom{4}{1} = 288$$ since there are three ways to choose the section from which two questions will be answered, $\binom{4}{2}$ ways to choose two questions from that section, and four ways to choose a single question from each of the other sections.
By the Inclusion-Exclusion Principle, there are $$\binom{12}{4} - \binom{3}{1}\binom{8}{4} + \binom{3}{2}\binom{4}{4} = 288$$ admissible ways to select the questions since there are $\binom{12}{4}$ ways to select four questions, three ways to exclude one section and $\binom{8}{4}$ ways to select four of the remaining eight questions, and $\binom{3}{2}$ ways to exclude two sections and $\binom{4}{4}$ ways to answer all four of the remaining four questions. The reason we must add the last term is that we subtracted each selection in which all the selected questions are drawn from one section twice, once for each way of designating one of the other two sections as the excluded section.
While the first method of solving this problem is simpler in this case, the Inclusion-Exclusion Principle can be more readily applied if the number of questions in each section ot the number of sections were large.