One, (50 points)
In an equilateral triangle $\triangle ABC$ with side length $a$, there is a moving point $P$ (which can be on the boundary). Find the maximum values of $PA \cdot PB$, $RC$, and $PA + PB \backslash PC$.

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$-,(50$ points $)$
First, find the maximum value of $P A+P B+P C$.
When $P B+P C$ is fixed, the trajectory of point $\boldsymbol{P}$ is an ellipse with foci at $B$ and $C$. By the convexity of the ellipse, $P A$ achieves its maximum value when $P$ is on the boundary.
Therefore, assume $P$ is on $A B$,
then $P A+P B+P C=A B+P C=u+P C \leqslant a+a=2 a$.
The maximum value is achieved when $P$ is at one of the vertices of $\triangle A B C$.
Next, find the maximum value of $P A \cdot P B \cdot P C$.
Establish a Cartesian coordinate system with $A$ as the origin (as shown in the figure),
point $B$ is at $(-1,-\sqrt{3})$, and point $C$ is at $(1,-\sqrt{3})$,
let $|PA| = a$, then point $P$ can be represented as $(a \cos \theta, a \sin \theta)$,
since $P$ is inside $\triangle A B C$, so $\theta \in\left[\frac{4}{3} \pi, \frac{5}{3} \pi\right]$,
i.e., $\sin \theta \in\left[-1, -\frac{\sqrt{3}}{2}\right]$,
$$
\begin{array}{l}
|P B| \cdot|P C|=\sqrt{(a \cos \theta+1)^{2}+(a \sin \theta+\sqrt{3})} \cdot \\
\sqrt{(a \cos \theta-1)^{2}+(a \sin \theta+\sqrt{3})^{2}} \\
=\sqrt{a^{2}+4+2 a \cos \theta+2 \sqrt{3} a \sin \theta} \cdot \sqrt{a^{2}+4+2 \sqrt{3} a \sin \theta-2 a \cos ^{2} \theta} \\
=\sqrt{\left(a^{2}+4+2 \sqrt{3} a \sin \theta\right)^{2}-4 a^{2} \cos ^{2} \theta} \\
=\sqrt{16 a^{2} \sin ^{2} \theta+4 \sqrt{3} a\left(a^{2}+4\right) \sin \theta+\left(a^{2}+4\right)^{2}-4 a^{2}},
\end{array}
$$

The maximum value of this expression can only be achieved when $\sin \theta=-1$ or $-\frac{\sqrt{3}}{2}$, i.e., on the boundary or on the altitude.
1. If on the boundary, assume it is on $B C$.
Let $A H$ be the altitude from $A$ to $B C$, and let $P H=b$,
then $A P=\sqrt{b^{2}+\left(\frac{\sqrt{3}}{2} a\right)^{2}}=\sqrt{b^{2}+\frac{3}{4} a^{2}}$,
$$
\begin{array}{l} 
B P \cdot P C=B H^{2}-P H^{2}=\frac{1}{4} a^{2}-b^{2}, \\
P A \cdot P B \cdot P C=\sqrt{b^{2}+\frac{3}{4} a^{2}} \cdot\left(\frac{1}{4} a^{2}-b^{2}\right) . \\
=\sqrt{b^{6}+\frac{1}{4} a^{2} b^{4}-\frac{5}{16} a^{4} b^{2}+\frac{3}{64} a^{6} .}
\end{array}
$$

Let $x=\frac{b^{2}}{a^{2}}, x \in\left[0, \frac{1}{4}\right]$,
then for $x^{3}+\frac{1}{4} x^{2}-\frac{5}{16} x+\frac{3}{64}$, the derivative is
$$
3 x^{2}+\frac{1}{2} x-\frac{5}{16}=\left(x-\frac{1}{4}\right)\left(3 x+\frac{5}{4}\right)<0 \text {. }
$$

So $x^{3}+\frac{1}{4} x^{2}-\frac{5}{16} x+\frac{3}{64}$ is monotonically decreasing in $\left[0, \frac{1}{4}\right]$, so the maximum value is achieved at $0$.
Thus, $P A \cdot P B \cdot P C=\sqrt{\frac{3}{64} a^{6}}=\frac{\sqrt{3}}{8} a^{2}$.
2. On the altitude, assume it is on $A H$.

Let $P H=c$, then $P B \cdot P C=P B^{2}=\left(\sqrt{P H^{2}+B H^{2}}\right)^{2}=$ $\left(\sqrt{c^{2}+\frac{1}{4} a^{2}}\right)^{2}=c^{2}+\frac{1}{4} a^{2}$.
$P A=\frac{\sqrt{3}}{2} a-c$, let $\frac{c}{a}=y, y \in[0,1]$.
$P A \cdot P B \cdot P C=a^{3} \cdot\left(-y^{3}+\frac{\sqrt{3}}{2} y^{2}-\frac{1}{4} y+\frac{\sqrt{3}}{8}\right) = \frac{\sqrt{3}}{8} a^{3}$.