3. (24 points) Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and let segment $P Q$ be a focal chord passing through the left focus $F$ and not perpendicular to the $x$-axis. If there exists a point $R$ on the left directrix such that $\triangle P Q R$ is an equilateral triangle, find the range of the eccentricity $e$ of the ellipse, and express the slope of line $P Q$ in terms of $e$.


3. As shown in Figure 4, let the midpoint of line segment $P Q$ be $M$, and draw perpendiculars from points $P$, $M$, and $Q$ to the directrix, with the feet of the perpendiculars being points $P$, $M^{\prime}$, and $Q^{\prime}$, respectively. Then
$$
\begin{array}{l}
\left|M M^{\prime}\right| \\
=\frac{1}{2}\left(\left|P P^{\prime}\right|+\left|Q Q^{\prime}\right|\right) \\
=\frac{1}{2}\left(\frac{|P F|}{e}+\frac{|Q F|}{e}\right)=\frac{|P Q|}{2 e} .
\end{array}
$$

Assume there exists a point $R$ such that $|R M|=\frac{\sqrt{3}}{2}|P Q|$, and $\left|M M^{\prime}\right|\frac{\sqrt{3}}{3}$. Therefore, $\cos \angle R M M^{\prime}=\frac{\left|M M^{\prime}\right|}{\left|R M^{\prime}\right|}=\frac{1}{2 e} \times \frac{2}{\sqrt{3}}=\frac{\sqrt{3}}{3 e}$,
$$
\cot \angle R M M^{\prime}=\frac{1}{\sqrt{3 e^{2}-1}} \text {. }
$$

In Figure 4, when $|P F|\frac{\sqrt{3}}{3}$, draw a focal chord $P Q$ through point $F$ with a slope of $\frac{1}{\sqrt{3 e^{2}-1}}$. Its perpendicular bisector intersects the left directrix at point $R$. From the above calculations, we know that $|R M|=\frac{\sqrt{3}}{2}|P Q|$. Therefore, $\triangle P Q R$ is an equilateral triangle.
By symmetry, when $|P F|>|Q F|$, we have
$$
k_{P Q}=-\frac{1}{\sqrt{3 e^{2}-1}} .
$$

Thus, the range of the eccentricity $e$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\left(\frac{\sqrt{3}}{3}, 1\right)$, and the slope of the line $P Q$ is $\pm \frac{1}{\sqrt{3 e^{2}-1}}$.