1. Given a regular tetrahedron $P-ABC$ with a volume of $9 \sqrt{3}$, the dihedral angle between the side face $PAB$ and the base $ABC$ is $60^{\circ}$. Point $D$ is on segment $AB$ such that $AD=\frac{1}{6} AB$, point $E$ is on segment $AC$ such that $AE=\frac{1}{6} AC$, and point $F$ is the midpoint of segment $PC$. Plane $DEF$ intersects segment $PB$ at point $G$. Find the area of quadrilateral $DEFG$.
Three, 1. Let the side length of the equilateral $\triangle ABC$ be $x$, the center be $O$, and the midpoint of line segment $BC$ be $M$. Connect $AM$, $PM$, and $PO$.

Since $AM \perp BC$ and $PM \perp BC$, we know that $\angle AMP$ is the plane angle of the dihedral angle formed by the lateral face and the base of the tetrahedron $P-ABC$.
Thus, $\angle AMP = 60^{\circ}$.
Therefore, $PO = OM \tan \angle AMP = \frac{1}{3} AM \tan 60^{\circ}$
$$
= \frac{1}{3} \times \frac{\sqrt{3}}{2} x \times \sqrt{3} = \frac{x}{2} \\
\Rightarrow V = \frac{1}{3} S_{\triangle ABC} \cdot PO = \frac{1}{3} \times \frac{\sqrt{3}}{4} x^2 \times \frac{x}{2} \\
= \frac{\sqrt{3}}{24} x^3 = 9 \sqrt{3} \\
\Rightarrow x = 6.
$$

Since $\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{6}$, we have
$DE \parallel BC$, $DE = \frac{1}{6} BC = 1$.
By $DE \parallel BC$, we know that $BC \parallel$ plane $DEF$.
Thus, $BC \parallel FG$.
Furthermore, since $F$ is the midpoint of line segment $PC$, we know that $G$ is the midpoint of $PB$, and $FG = \frac{1}{2} BC = 3$.
By $BD = CE$, $BG = CF$, and $\angle ABP = \angle ACP$
$\Rightarrow \triangle BDG \cong \triangle CEF \Rightarrow DG = EF$.
By $DE \parallel BC \parallel FG$, we know that quadrilateral $DEFG$ is an isosceles trapezoid with bases $DE = 1$ and $FG = 3$.

Let the midpoint of $DE$ be $K$ and the midpoint of $FG$ be $L$. Then $KL$ is the height of the isosceles trapezoid $DEFG$.
Thus, point $K$ lies on $AM$, and $AK = \frac{1}{6} AM$, so
$$
KM = \frac{5}{6} AM = \frac{5}{6} \times 3 \sqrt{3} = \frac{5 \sqrt{3}}{2} ;
$$

Point $L$ lies on $PM$, and $PL = \frac{1}{2} PM$, so
$$
LM = \frac{1}{2} PM = OM = \frac{1}{3} AM = \frac{1}{3} \times 3 \sqrt{3} = \sqrt{3} .
$$

In $\triangle KLM$, by the cosine rule, we have
$$
\begin{array}{l}
KL = \sqrt{KM^2 + LM^2 - 2 KM \cdot LM \cos 60^{\circ}} \\
= \sqrt{\frac{75}{4} + 3 - \frac{15}{2}} = \frac{\sqrt{57}}{2} .
\end{array}
$$

Thus, $S_{\text{trapezoid } DEFG} = \frac{(DE + FG) KL}{2} = \sqrt{57}$.