3. Given the set $A=\left\{x \mid x^{2}-5 x+6<0\right\}$, $B=\left\{\left.x\right| x^{2}-a x+5=0\right\}$. If $A \cap B \neq \varnothing$, then the range of real number $a$ is ( ).
(A) $\left[2 \sqrt{5}, \frac{14}{3}\right)$
(B) $\left[2 \sqrt{5}, \frac{9}{2}\right)$
(C) $\left(\frac{9}{2}, \frac{14}{3}\right)$
(D) $\left(-\frac{14}{3},-2 \sqrt{5}\right)$
3.A.

Since $A=\{x \mid 2<x<3\}$, and $A \cap B \neq \varnothing$, the equation $x^{2}-a x+5=0$ has a solution in $(2,3)$. The problem is converted to finding the range of $a=\frac{x^{2}+5}{x}=x+\frac{5}{x}$ in $(2,3)$.

Since $x+\frac{5}{x}$ is monotonically decreasing on $(2, \sqrt{5}]$ and monotonically increasing on $[\sqrt{5}, 3)$, the range of $x+\frac{5}{x}$ is $\left[2 \sqrt{5}, \frac{14}{3}\right)$. Therefore, the range of the real number $a$ is $\left[2 \sqrt{5}, \frac{14}{3}\right)$.