12. As shown in Figure 4, let points $A(-\sqrt{5}, 0)$ and $B(\sqrt{5}, 0)$, and the incenter of $\triangle ABC$ moves on the line $x=2$.
(1) Find the equation of the locus of point $C$;
(2) If two rays are drawn from point
$M(2,0)$, intersecting the locus found in (1)
at points
$P$ and $Q$, and $\overrightarrow{M P} \cdot \overrightarrow{M Q}$
$=0$, prove that the line $PQ$ must pass through a fixed point.
12. (1) Let the incircle of $\triangle ABC$ touch side $AB$ at point $D$. Then
$$
\begin{array}{l}
|CA| - |CB| = |AD| - |BD| \\
= (\sqrt{5} + 2) - (\sqrt{5} - 2) = 4
\end{array}
$$
(2) Let $l_{PQ}: x = my + a (a > 2)$. Substituting into $\frac{x^2}{4} - y^2 = 1$, we get
$$
\left(m^2 - 4\right) y^2 + 2am y + a^2 - 4 = 0.
$$

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$. Then
$$
\begin{array}{l}
y_1 + y_2 = -\frac{2am}{m^2 - 4}, \quad y_1 y_2 = \frac{a^2 - 4}{m^2 - 4}. \\
\text{Also, } \overrightarrow{MP} \cdot \overrightarrow{MQ} = (x_1 - 2)(x_2 - 2) + y_1 y_2 \\
= (my_1 + a - 2)(my_2 + a - 2) + y_1 y_2 \\
= (m^2 + 1) y_1 y_2 + m(a - 2)(y_1 + y_2) + (a - 2)^2 \\
= 0, \\
\text{Thus, } \frac{(m^2 + 1)(a^2 - 4)}{m^2 - 4} - \frac{2am^2(a - 2)}{m^2 - 4} + (a - 2)^2 = 0.
\end{array}
$$

Simplifying, we get $3a^2 - 16a + 20 = 0$.
Solving, we get $a = 2$ (discard) or $a = \frac{10}{3}$.
Therefore, the line $PQ$ must pass through the fixed point $\left(\frac{10}{3}, 0\right)$.