Example 3 In $\triangle ABC$ with a fixed perimeter, it is known that $AB=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
(1) Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ intersecting the curve from (1) at points $M$ and $N$, and find the minimum value of $|\overrightarrow{B M}||\overrightarrow{B N}|$.
(1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of segment $AB$ as the $y$-axis.

Let $|CA| + |CB| = 2a (a > 3)$ be a constant. Then the locus of point $C$ is an ellipse with foci at $A$ and $B$.
Thus, the focal distance $2c = |AB| = 6$.
Notice,
$$
\begin{array}{l}
\cos C = \frac{|CA|^2 + |CB|^2 - |AB|^2}{2|CA||CB|} \\
= \frac{(|CA| + |CB|)^2 - 2|CA||CB| - |AB|^2}{2|CA||CB|} \\
= \frac{2a^2 - 18}{|CA||CB|} - 1.
\end{array}
$$

Also, $|CA||CB| \leq \left(\frac{|CA| + |CB|}{2}\right)^2 = a^2$, so $\cos C \geq 1 - \frac{18}{a^2}$.
From the problem, $1 - \frac{18}{a^2} = \frac{7}{25} \Rightarrow a^2 = 25$.
At this point, $|PA| = |PB|$, and $P(0, \pm 4)$.
Therefore, the equation of the locus of point $C$ is
$$
\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0).
$$
(2) Establish a polar coordinate system with $A$ as the pole and $Ax$ as the polar axis. Then the polar equation of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0)$ is
$$
\rho = \frac{16}{5 - 3 \cos \theta} (\theta \neq 0 \text{ and } \theta \neq \pi).
$$

Let the inclination angle of line $MN$ be $\theta$. Then
$$
|AM| = \frac{16}{5 - 3 \cos \theta}, |BM| = \frac{16}{5 + 3 \cos \theta}.
$$

By the definition of the ellipse,
$$
\begin{array}{l}
|\overrightarrow{BM}||\overrightarrow{BN}| = (10 - |AM|)(10 - |BM|) \\
= 100 - 10\left(\frac{16}{5 - 3 \cos \theta} + \frac{16}{5 + 3 \cos \theta}\right) + \\
\frac{16}{5 - 3 \cos \theta} \cdot \frac{16}{5 + 3 \cos \theta} \\
= 100 - \frac{16 \times 84}{25 - 9 \cos^2 \theta}.
\end{array}
$$

If there are no restrictions, when $\cos^2 \theta = 1$, $|\overrightarrow{BM}||\overrightarrow{BN}|$ achieves its minimum value of 16.

However, $0 < \theta < \pi$, so such $M$ and $N$ do not exist, meaning the set of minimum values of $|\overrightarrow{BM}||\overrightarrow{BN}|$ is empty.