3. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=B C=4, A A_{1}=2, P$ is a point on the plane $A_{1} B C$, and $\overrightarrow{D P} \cdot \overrightarrow{P B}=0$. Then, the area of the plane region enclosed by all points $P$ that satisfy the above conditions is $\qquad$.
Given, the midpoint of $DB$ is $O$, and $\angle D P B=90^{\circ}$
$\Rightarrow O P=O B=O D=2 \sqrt{2}$, which means point $P$ lies on a sphere with
$O$ as the center and $2 \sqrt{2}$ as the radius.
It is easy to see that $B C \perp$ plane $E G F \Rightarrow$ plane $A_{1} B C \perp$
plane $E G F \Rightarrow O O_{1} \perp$ plane $A_{1} B C$,
and $\triangle O O_{1} F \sim \triangle E G F \Rightarrow \frac{O O_{1}}{E G}=\frac{O F}{E F}$
$\Rightarrow O O_{1}=\frac{E G \cdot O F}{E F}=\frac{2 \times 2}{2 \sqrt{5}}=\frac{2}{\sqrt{5}}$, thus $O_{1} P=\sqrt{8-\frac{4}{5}}=\frac{6}{\sqrt{5}}$, meaning point $P$ traces a circle with $O_{1}$ as the center and $\frac{6}{\sqrt{5}}$ as the radius. Therefore, the area of the plane region enclosed by point $P$ is $\frac{36 \pi}{5}$.