2. Let the height of the regular triangular prism $A B C-A_{1} B_{1} C_{1}$ be 2, and the side length of the base be 1. The centroid of the upper base $\triangle A_{1} B_{1} C_{1}$ is $P$. A plane $B C D \perp A P$ is made through the lower base edge $B C$, intersecting the edge $A A_{1}$ at point $D$. Then the area of the section $\triangle B C D$ is $\qquad$
Prompt As shown in Figure 12, establish a spatial rectangular coordinate system.
Let $A D=h$. Then
$A(0,0,0)$,
$B(1,0,0)$,
$C\left(\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\right)$,
$A_{1}(0,0,2)$,
$B_{1}(1,0,2)$,
$C_{1}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}, 2\right)$,
$P\left(\frac{1}{2},-\frac{\sqrt{3}}{6}, 2\right), D(0,0, h)$.
Since $A P \perp$ plane $B C D$,
$\overrightarrow{A P}=\left(\frac{1}{2},-\frac{\sqrt{3}}{6}, 2\right), \overrightarrow{B D}=(-1,0, h)$,
then $\overrightarrow{A P} \cdot \overrightarrow{B D}=0 \Rightarrow h=\frac{1}{4} \Rightarrow D\left(0,0, \frac{1}{4}\right)$.
Also, since $B C=1$,
$$
\begin{array}{l}
C D=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{4}\right)^{2}}=\frac{\sqrt{17}}{4}, \\
B D=\sqrt{1+\left(\frac{1}{4}\right)^{2}}=\frac{\sqrt{17}}{4},
\end{array}
$$

Therefore, the semi-perimeter of $\triangle B C D$ is $p=\frac{1}{2}+\frac{\sqrt{17}}{4}$.
By Heron's formula,
$$
S=\sqrt{p(p-a)(p-b)(p-c)}=\frac{\sqrt{13}}{8} .
$$