3. Given $M=\left\{x \left\lvert\, \frac{a x-6}{x^{2}-a}<0\right.\right\}$, and $\left\{\begin{array}{l}2 \in M, \\ 3 \notin M .\end{array}\right.$ Then the range of values for $\boldsymbol{a}$ is $\qquad$ .
3. $[2,3) \cup(4,9]$.

From $2 \in M$, we have $\frac{2a-6}{2^2-a}<0$.
Solving this, we get $a \in (-\infty, 3) \cup (4, +\infty)$.
Also, from $3 \notin M$, we have $\frac{3a-6}{3^2-a} \geqslant 0$ or $a=9$.
Solving this, we get $a \in [2,9]$.
Combining these, we get $[2,3) \cup (4,9]$.