14. Given a regular tetrahedron $S-ABC$ with height $SO=3$, and the side length of the base is 6. Draw a perpendicular from point $A$ to the opposite face $SBC$, and let the foot of the perpendicular be $D$. On $AD$, take a point $P$ such that $\frac{AP}{PD}=8$. Find the area of the section passing through point $P$ and parallel to the base.
Let the midpoint of $BC$ be $E$. Since $S-ABC$ is a regular tetrahedron, points $A, O, E$ are collinear, and points $S, D, E$ are collinear.
$$
\begin{array}{l}
SO = BE = 3 \Rightarrow SE = AO = 2\sqrt{3} \\
\Rightarrow SA = \sqrt{21} \Rightarrow \cos \angle SEA = \frac{12 + 27 - 21}{2 \cdot 2\sqrt{3} \cdot 3\sqrt{3}} \\
= \frac{1}{2} \Rightarrow \angle SEA = 60^{\circ} \Rightarrow \angle EAD = \angle ESO = 30^{\circ}
\end{array}
$$
$\Rightarrow AD = \frac{9}{2}, ED = \frac{3\sqrt{3}}{2} \Rightarrow SD = \frac{\sqrt{3}}{2}$. Let the intersection of $AD$ and $SO$ be point $P'$. Thus, $P'D = \frac{1}{2} \Rightarrow \frac{AP'}{P'D} = 8 \Rightarrow P$ coincides with $P' \Rightarrow \frac{SP}{SO} = \frac{1}{3}$. Let the area of the section through $P$ and parallel to the base be $S \Rightarrow \frac{S}{S_{\triangle ABC}} = \frac{1}{9} \Rightarrow S = \sqrt{3}$.