3. Let $F_{1}, F_{2}$ be the left and right foci of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, respectively. A line passing through point $F_{1}$ intersects the ellipse $E$ at points $A$ and $B$, with $\left|A F_{1}\right|=3\left|B F_{1}\right|$ and $\cos \angle A F_{2} B=\frac{3}{5}$. As shown in the figure, the line $y=\frac{1}{2} x$ intersects the ellipse $E$ at points $P$ and $Q$. Points $C$ and $D$ are on the ellipse $E$ and are different from $P$ and $Q$. The lines $P C$ and $Q D$ intersect at point $M$, and the lines $P D$ and $Q C$ intersect at point $N$. Then the slope of the line $M N$ is $\qquad$.
3. -1

Analysis: Let $\left|F_{1} B\right|=k$, then $k>0$ and $\left|A F_{1}\right|=3 k,|A B|=4 k$. By the definition of an ellipse, we have $\left|A F_{2}\right|=2 a-3 k,\left|B F_{2}\right|=2 a-k$. In $\triangle A B F_{2}$, by the cosine rule, we get $|A B|^{2}=\left|A F_{2}\right|^{2}+\left|B F_{2}\right|^{2}-2\left|A F_{2}\right| \cdot\left|B F_{2}\right| \cos \angle A F_{2} B$, which is $(4 k)^{2}=(2 a-3 k)^{2}+(2 a-k)^{2}-\frac{6}{5}(2 a-3 k) \cdot(2 a-k)$. Simplifying, we get: $\quad(a+k)(a-3 k)=0$, hence $a=3 k$. Therefore, $\left|B F_{2}\right|^{2}=\left|F_{2} A\right|^{2}+|A B|^{2}$, which implies $F_{1} A \perp F_{2} A$, meaning $\triangle A F_{1} F_{2}$ is an isosceles right triangle. Thus, $a=\sqrt{2} b$. Let $P\left(x_{0}, y_{0}\right), M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, then $Q\left(-x_{0},-y_{0}\right)$.

It is easy to get $k_{C P} k_{C Q}=-\frac{b^{2}}{a^{2}}=-\frac{1}{2}$, which means $k_{M P} k_{N Q}=-\frac{b^{2}}{a^{2}}=-\frac{1}{2}$. Similarly, $k_{N P} k_{M Q}=-\frac{1}{2}$
which gives $\left\{\begin{array}{l}\frac{y_{1}-y_{0}}{x_{1}-x_{0}} \times \frac{y_{2}+y_{0}}{x_{2}+x_{0}}=-\frac{1}{2} \\ \frac{y_{1}+y_{0}}{x_{1}+x_{0}} \times \frac{y_{2}-y_{0}}{x_{2}-x_{0}}=-\frac{1}{2}\end{array} \Rightarrow\left\{\begin{array}{l}y_{1} y_{2}-y_{0}\left(y_{2}-y_{1}\right)-y_{0}^{2}=-\frac{1}{2}\left[x_{1} x_{2}-x_{0}\left(x_{2}-x_{1}\right)-x_{0}^{2}\right] \\ y_{1} y_{2}+y_{0}\left(y_{2}-y_{1}\right)-y_{0}^{2}=-\frac{1}{2}\left[x_{1} x_{2}+x_{0}\left(x_{2}-x_{1}\right)-x_{0}^{2}\right]\end{array}\right.\right.$,
subtracting the two equations, we get $2 y_{0}\left(y_{2}-y_{1}\right)=-\frac{1}{2} \cdot 2 x_{0}\left(x_{2}-x_{1}\right)$, so $k_{M N}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=-\frac{1}{2} \cdot \frac{x_{0}}{y_{0}}=-1$.