Example 1 (1998 National High School League Question) If the non-empty set $A=\{x \mid 2 a+1 \leqslant x \leqslant 3 a-5\}, B=$ $\{x \mid 3 \leqslant x \leqslant 32\}$, then the set of all $a$ that makes $A \subseteq A \cap B$ true is ( ).
A. $\{a \mid 1 \leqslant a \leqslant 9\}$
B. $\{a \mid 6 \leqslant a \leqslant 9\}$
C. $\{a \mid a \leqslant 9\}$
D. $\varnothing$
Solution: Choose B. Reason: According to the conditions of the problem, we get $A \subseteq B$, which is equivalent to
$$
\left\{\begin{array}{l}
2 a+1 \geqslant 3, \\
3 a-5 \leqslant 32, \\
3 a-5 \geqslant 2 a+1,
\end{array}\right.
$$

Solving these inequalities, we get $6 \leqslant a \leqslant 9$, hence the answer is B.