Three, (15 points) In the Cartesian coordinate plane, $\triangle ABC$ has two vertices $A, B$ with coordinates $A(-1,0)$, $B(1,0)$. Two points $G, M$ in the plane simultaneously satisfy the following conditions:
(1) $\boldsymbol{G A}+\boldsymbol{G B}+\boldsymbol{G C}=0$;
(2) $|M A|=|M B|=|M C|$;
(3) $G M \parallel A B$.
(1) Find the equation of the locus of vertex $C$ of $\triangle ABC$;
(2) A line $l$ passing through point $P(3,0)$ intersects the locus from (1) at points $E, F$. Find the range of $\boldsymbol{P E} \cdot \boldsymbol{P F}$.
(1) Let $C(x, y)$, $G(x_0, y_0)$, and $M(x_M, y_M)$.
Since $|MA| = |MB|$, point $M$ lies on the perpendicular bisector of segment $AB$.
Given $A(-1,0)$ and $B(1,0)$, we have $x_M = 0$.
Since $GM \parallel AB$, we have $y_M = y_0$.
Also, $GA + GB + GC = 0$, which means
$$
\begin{array}{l}
\left(-1 - x_0, -y_0\right) + \left(1 - x_0, -y_0\right) + \left(x - x_0, y - y_0\right) \\
= (0, 0).
\end{array}
$$

Solving this, we get $x_0 = \frac{x}{3}$ and $y_0 = \frac{y}{3}$. Thus, $y_M = \frac{y}{3}$.
Since $|MB| = |MC|$, we have
$$
\sqrt{(0 - 1)^2 + \left(\frac{y}{3} - 0\right)^2} = \sqrt{(0 - x)^2 + \left(\frac{y}{3} - y\right)^2}.
$$

This simplifies to $x^2 + \frac{y^2}{3} = 1$ (with $y \neq 0$).
Therefore, the locus equation of point $C$ is $x^2 + \frac{y^2}{3} = 1$ (with $y \neq 0$).
(2) Let the equation of line $l$ be $y = k(x - 3)$.

Suppose $E(x_1, y_1)$ and $F(x_2, y_2)$. From
$$
\left\{\begin{array}{l}
y = k(x - 3), \\
x^2 + \frac{y^2}{3} = 1,
\end{array}\right.
$$

eliminating $y$ gives
$$
\left(k^2 + 3\right) x^2 - 6k^2 x + 9k^2 - 3 = 0.
$$

Thus, $x_1 + x_2 = \frac{6k^2}{k^2 + 3}$ and $x_1 x_2 = \frac{9k^2 - 3}{k^2 + 3}$.
And $\boldsymbol{PE} \cdot \boldsymbol{PF} = |PE| \cdot |PF| \cdot \cos 0^\circ = |PE| \cdot |PF|$
$$
\begin{array}{l}
= \sqrt{1 + k^2} |3 - x_1| \cdot \sqrt{1 + k^2} |3 - x_2| \\
= (1 + k^2) |9 - 3(x_1 + x_2) + x_1 x_2| \\
= (1 + k^2) \left| \frac{9k^2 + 27 - 18k^2 + 9k^2 - 3}{k^2 + 3} \right| \\
= \frac{24(k^2 + 1)}{k^2 + 3} = 24 - \frac{48}{k^2 + 3}.
\end{array}
$$

From equation (1), we have
$$
\Delta = (6k^2)^2 - 4(k^2 + 3)(9k^2 - 3) > 0.
$$

Solving this, we get $k^2 < \frac{3}{8}$.
Since $k \neq 0$, we have $0 < k^2 < \frac{3}{8}$.
Thus, $k^2 + 3 \in \left(3, \frac{27}{8}\right)$.
Therefore, $PE \cdot PF \in \left(8, \frac{88}{9}\right)$.