3. Given an equilateral triangle $ABC$ with side length 4, points $D$, $E$, and $F$ are on $BC$, $CA$, and $AB$ respectively, and $|AE|=|BF|=|CD|=1$. Connect $AD$, $BE$, and $CF$, intersecting to form $\triangle RQS$. Point $P$ moves within $\triangle RQS$ and on its sides, and the distances from $P$ to the three sides of $\triangle ABC$ are denoted as $x$, $y$, and $z$.
(1) Prove that when point $P$ is at the vertices of $\triangle RQS$, the product $xyz$ has a minimum value;
(2) Find the minimum value of the above $xyz$.
(1) When point $P$ moves on line $GH$, $z$ is fixed,

Since $x+y+z=|CN|=2\sqrt{3}$,
and $xy=\frac{(x+y)^2-(x-y)^2}{4}$, then when point $P$ falls on
the side of $\triangle RQS$, $|x-y|$ reaches its maximum, thus $xy$ is
the smallest. Below, we only consider the case where point $P$ moves on the three sides of $\triangle RQS$.
When point $P$ moves on side $SQ$, in $\triangle CFB$, applying Menelaus' theorem to line $ASD$ gives $\frac{CS}{SF} \cdot \frac{FA}{AB} \cdot \frac{BD}{DC}=1 \Rightarrow \frac{CS}{SF}=\frac{4}{9}$, in $\triangle CAF$, applying Menelaus' theorem to line $BQE$ gives $\frac{CE}{EA} \cdot \frac{AB}{BF} \cdot \frac{FQ}{QC}=1 \Rightarrow \frac{FQ}{QC}=\frac{1}{12} \Rightarrow z \in\left[\frac{2\sqrt{3}}{13}, \frac{18\sqrt{3}}{13}\right]$.
By $\frac{PH}{FB}=1-\frac{z}{2\sqrt{3}} \Rightarrow |PH|=1-\frac{z}{2\sqrt{3}} \Rightarrow x=\frac{\sqrt{3}}{2}|PH|=\frac{\sqrt{3}}{2}-\frac{1}{4}z$,
$|PG|=3|PH| \Rightarrow y=\frac{3\sqrt{3}}{2}-\frac{3}{4}z \Rightarrow xyz=\frac{3}{16}z(2\sqrt{3}-z)^2$.
Let $f(z)=\frac{3}{16}z(2\sqrt{3}-z)^2, z \in\left[\frac{2\sqrt{3}}{13}, \frac{18\sqrt{3}}{13}\right], f'(z)=\frac{3}{16}(2\sqrt{3}-z)(2\sqrt{3}-3z)$,
so $f(x)$ is monotonically increasing on $\left(\frac{2\sqrt{3}}{13}, \frac{2\sqrt{3}}{3}\right)$ and monotonically decreasing on $\left(\frac{2\sqrt{3}}{3}, \frac{18\sqrt{3}}{13}\right)$,
and $f\left(\frac{2\sqrt{3}}{13}\right)=\frac{3}{16} \cdot \frac{2\sqrt{3}}{13} \cdot\left(\frac{24\sqrt{3}}{13}\right)^2=\frac{648\sqrt{3}}{2197}, f\left(\frac{18\sqrt{3}}{13}\right)=\frac{648\sqrt{3}}{2197}$.
Thus, when point $P$ coincides with $Q, S$, the product $xyz$ has a minimum value. Similarly, the other cases can be proven.
In summary, when point $P$ is at the vertices of $\triangle RQS$, the product $xyz$ has a minimum value.
(2) From (1), the minimum value of $xyz$ is $\frac{648\sqrt{3}}{2197}$.