Three. (20 points) In the Cartesian coordinate system $x O y$, it is known that the ellipse $C: \frac{x^{2}}{27}+\frac{y^{2}}{18}=1$ has its right focus at $F$. A line $l$ passing through point $F$ intersects the ellipse $C$ at points $A$ and $B$. Is there a fixed point $P$ on the $x$-axis such that when the line $l$ rotates around point $F$, $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is always a constant?
Three, from the problem, we know the point $F(3,0)$.
Let point $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
When the line $l$ is not perpendicular to the $x$-axis, let the equation of $l$ be $y=k(x-3)$.
Substituting into the ellipse equation, we get
$$
\begin{array}{l}
\left(2+3 k^{2}\right) x^{2}-18 k^{2} x+27 k^{2}-54=0 \\
\Rightarrow x_{1}+x_{2}=\frac{18 k^{2}}{2+3 k^{2}}, x_{1} x_{2}=\frac{27 k^{2}-54}{2+3 k^{2}} .
\end{array}
$$

Assume there exists a fixed point $P(t, 0)$ on the $x$-axis such that $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is a constant. Then
$$
\begin{array}{l}
\overrightarrow{P A} \cdot \overrightarrow{P B}=\left(x_{1}-t, y_{1}\right) \cdot\left(x_{2}-t, y_{2}\right) \\
=x_{1} x_{2}-t\left(x_{1}+x_{2}\right)+t^{2}+y_{1} y_{2} \\
=x_{1} x_{2}-t\left(x_{1}+x_{2}\right)+t^{2}+k\left(x_{1}-3\right) k\left(x_{2}-3\right) \\
=\left(1+k^{2}\right) x_{1} x_{2}-\left(3 k^{2}+t\right)\left(x_{1}+x_{2}\right)+t^{2}+9 k^{2} \\
=\left(1+k^{2}\right) \frac{27 k^{2}-54}{2+3 k^{2}}-\left(3 k^{2}+t\right) \frac{18 k^{2}}{2+3 k^{2}}+t^{2}+9 k^{2} \\
=-\frac{54+(18 t+9) k^{2}}{2+3 k^{2}}+t^{2} \text {. } \\
\end{array}
$$

When the line $l$ rotates around point $F$, i.e., $k$ changes, to make $\overrightarrow{P A} \cdot \overrightarrow{P B}$ a constant, $\frac{54+(18 t+9) k^{2}}{2+3 k^{2}}$ must be a constant.
Thus, $\frac{54}{2}=\frac{18 t+9}{3} \Rightarrow t=4$.
At this point, $\overrightarrow{P A} \cdot \overrightarrow{P B}=-11$.
When the line $l$ is perpendicular to the $x$-axis,
$A(3,2 \sqrt{3}), B(3,-2 \sqrt{3})$.
Hence, $\overrightarrow{P A} \cdot \overrightarrow{P B}=(3-4,2 \sqrt{3}) \cdot(3-4,-2 \sqrt{3})=-11$.
In summary, there exists a fixed point $P(4,0)$ on the $x$-axis such that $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is a constant.