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 $\qquad$
(2016, National High School Mathematics League Shanxi Province Preliminary)
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As shown in Figure 12, establish a spatial rectangular coordinate system.
\[
\begin{array}{c}
\text { Let } A D=h \text {. Then } \\
A(0,0,0), \\
B(1,0,0), \\
C\left(\frac{1}{2},-\frac{\sqrt{3}}{2}, 0\right), \\
A_{1}(0,0,2), \\
B_{1}(1,0,2), \\
C_{1}\left(\frac{1}{2},-\frac{\sqrt{3}}{2}, 2\right), \\
P\left(\frac{1}{2},-\frac{\sqrt{3}}{6}, 2\right), D(0,0, h) .
\end{array}
\]

Since \( A P \perp \) plane \( B C D \),
\[
\begin{array}{l}
\overrightarrow{A P}=\left(\frac{1}{2},-\frac{\sqrt{3}}{6}, 2\right), \overrightarrow{B D}=(-1,0, h), \\
\text { then } \overrightarrow{A P} \cdot \overrightarrow{B D}=0 \Rightarrow h=\frac{1}{4} \Rightarrow D\left(0,0, \frac{1}{4}\right) .
\end{array}
\]

Since \( B C=1 \),
\[
\begin{array}{l}
C D=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{4}\right)^{2}}=\frac{\sqrt{17}}{4}, \\
B D=\sqrt{1+\left(\frac{1}{4}\right)^{2}}=\frac{\sqrt{17}}{4},
\end{array}
\]

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