3. Given sets
$$
\begin{array}{c}
A=\left\{x \mid 2 x^{3}+5 x^{2}+x-2>0\right\}, \\
B=\left\{x \mid x^{2}-a x+b \leqslant 0\right\} .
\end{array}
$$

If sets $A$ and $B$ satisfy
$$
A \cap B=\left\{x \left\lvert\, \frac{1}{2}<x \leqslant 3\right.\right\},
$$

then the trajectory equation of point $(a, b)$ in the rectangular coordinate plane $a O b$ is . $\qquad$
3. $b=3 a-9\left(2 \leqslant a \leqslant \frac{7}{2}\right)$.

From $2 x^{3}+5 x^{2}+x-2>0$
$\Rightarrow(x+1)(x+2)(2 x-1)>0$
$\Rightarrow-2<x<-1$ or $x>\frac{1}{2}$
$\Rightarrow A=(-2,1) \cup\left(\frac{1}{2},+\infty\right)$.
Let $B=\{x \mid \alpha \leqslant x \leqslant \beta\}$.
Given $A \cap B=\left\{x \left\lvert\, \frac{1}{2}<x \leqslant 3\right.\right\}$, we know
$\beta=3,-1 \leqslant \alpha \leqslant \frac{1}{2}$.
By Vieta's formulas, we have $\alpha+\beta=a, \alpha \beta=b$.
Substituting $\beta=3$ to eliminate $\alpha$, we get
$$
b=3 a-9\left(2 \leqslant a \leqslant \frac{7}{2}\right) \text {, }
$$

This is the equation of the trajectory of the point $(a, b)$ in the rectangular coordinate plane $a O b$.