3. 0 .

Draw $AC \perp y$-axis at point $C$, and $AG \perp l$ at point $G$. Let $F$ be the right focus of the ellipse $\Gamma_{1}$.

Assume point $A$ is above the $x$-axis, and the focus of the parabola $\Gamma_{2}$ is $F$, with the directrix being the $y$-axis.
Thus, $AF = AC$.
$$
\text{By } \frac{AG}{AF}=\frac{a}{c}, \text{ we know } \frac{AG}{AC}=\frac{a}{c} \text{. }
$$

Since $OC \parallel DE$, we have
$$
\frac{AE}{AO}=\frac{AG}{AC}=\frac{a}{c} \text{. }
$$

Extend $BA$ to intersect $DE$ at point $H$.
$$
\text{Then } \frac{S_{\triangle ABE}}{S_{\triangle ABD}}=\frac{HE}{HD}=\frac{a}{c}=\frac{AE}{AO} \text{. }
$$

Therefore, $AG \parallel OD$, which means $AB \parallel x$-axis.
Thus, the slope of $AB$ is 0.
3. Given the ellipse $\Gamma_{1}: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $c=\sqrt{a^{2}-b^{2}}$, the parabola $\Gamma_{2}: y^{2}=2 c\left(x-\frac{c}{2}\right)$ intersects with $\Gamma_{1}$, and one of the intersection points is $A$. Let $O$ be the origin. The line $l: x=\frac{a^{2}}{c}$ intersects the $x$-axis and the line $O A$ at points $D$ and $E$, respectively. A point $B$ on the ellipse $\Gamma_{1}$ satisfies $\frac{S_{\triangle A B E}}{S_{\triangle A B D}}=\frac{a}{c}$, and points $A$ and $B$ are on opposite sides of the $y$-axis. Then the slope of $A B$ is . $\qquad$