He can choose the following combinations of questions from each section:
$(2,2,1);(2,1,2);(1,2,2);(3,1,1);(1,3,1);(1,1,3)$
Each of the first three combinations has $\binom{5}{2}\cdot \binom{5}{2}\cdot \binom{5}{1}$ ways. And each of the next three combinations has $\binom{5}{3}\cdot \binom{5}{1}\cdot \binom{5}{1}$ ways.
In total he can make $3\cdot \left(\binom{5}{2}\cdot \binom{5}{2}\cdot \binom{5}{1}+ \binom{5}{3}\cdot \binom{5}{1}\cdot \binom{5}{1}\right)=2250$ choices.
For every section we have one path. We choose 3 question from each section. For every path we have to add ($\text{not multiply}$) the ways
$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)$
Now we make a case decision.
a) We choose 2 questions from one (other) section and 2 questions from the remaining section.
$ \binom{5}{2}\cdot \binom{5}{2}$
b) We choose 1 questions from one (other) section and 3 questions from the remaining section.
$ \binom{5}{1}\cdot \binom{5}{3}$
Therefore in total we have
$\left( \binom{5}{1}+\binom{5}{1}+\binom{5}{1}\right)\cdot \left(\binom{5}{2}\cdot \binom{5}{2}+ \binom{5}{1}\cdot \binom{5}{3}\right)=2250$
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
Your way of counting is correct if you are required to mark one of your selected questions from each section with a star. But this overcounts because we don't care which question is your "starred" representative of the section.
For simplicity, suppose the 15 questions are numbered 1 to 15, with the first 5 questions being in the first section, and so on.
One outcome is to choose questions 1, 2, 3, 6, 11. However, you count this three times:
Another outcome is to choose questions 1, 2, 6, 7, 11. However, you count this four times:
In general for each actual outcome, you overcount by either 3 times or 4 times, which explains why $8250/2250 = 3.\bar{6}$ is between 3 and 4.