11. In rectangle $A B C D$, $A B=3, A D=4, P$ is a point on the plane of rectangle $A B C D$, satisfying $P A=2$, $P C=\sqrt{21}$. Then $\overrightarrow{P B} \cdot \overrightarrow{P D}=$
11.0.

As shown in Figure 3, let $A C$ and $B D$ intersect at point $E$, and connect $P E$. Then $E$ is the midpoint of $A C$ and $B D$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{P B} \cdot \overrightarrow{P D}=\frac{1}{4}\left[(\overrightarrow{P B}+\overrightarrow{P D})^{2}-(\overrightarrow{P B}-\overrightarrow{P D})^{2}\right] \\
=\frac{1}{4}\left[(2 \overrightarrow{P E})^{2}-(\overrightarrow{D B})^{2}\right] \\
=|\overrightarrow{P E}|^{2}-\frac{1}{4}|\overrightarrow{D B}|^{2} .
\end{array}
$$

Similarly, $\overrightarrow{P A} \cdot \overrightarrow{P C}=|\overrightarrow{P E}|^{2}-\frac{1}{4}|\overrightarrow{C A}|^{2}$.
Since $|\overrightarrow{D B}|=|\overrightarrow{C A}|$, we have,
$$
\overrightarrow{P B} \cdot \overrightarrow{P D}=\overrightarrow{P A} \cdot \overrightarrow{P C}
$$

Given $A C=B D=\sqrt{3^{2}+4^{2}}=5=\sqrt{P A^{2}+P C^{2}}$
$$
\Rightarrow \angle A P C=90^{\circ} \text {. }
$$

Therefore, $\overrightarrow{P B} \cdot \overrightarrow{P D}=\overrightarrow{P A} \cdot \overrightarrow{P C}=0$.