(12) As shown in the figure, given two points $A(-\sqrt{5}, 0)$ and $B(\sqrt{5}, 0)$, the incenter of $\triangle ABC$ moves on the line $x=2$.
(1) Find the equation of the locus of point $C$;
(2) Draw two rays from point $M(2,0)$, intersecting the locus found in (1) at points $P$ and $Q$, respectively, 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
$$
\begin{array}{l}
|C A|-|C B|=|A D|-|B D| \\
=(\sqrt{5}+2)-(\sqrt{5}-2)=42).
\end{array}
$$
(2) Let $P Q: x=m y+a(a>2)$, substitute into $\frac{x^{2}}{4}-y^{2}=1$, we get
$$
\left(m^{2}-4\right) y^{2}+2 a m y+a^{2}-4=0 \text {. }
$$

Let $P\left(x_{1}, y_{1}\right)$ and $Q\left(x_{2}, y_{2}\right)$, then $y_{1}+y_{2}=-\frac{2 a m}{m^{2}-4}$,
$$
y_{1} y_{2}=\frac{a^{2}-4}{m^{2}-4} \text {. }
$$

Since
$$
\begin{aligned}
\overrightarrow{M P} \cdot \overrightarrow{M Q} & =\left(x_{1}-2\right)\left(x_{2}-2\right)+y_{1} y_{2} \\
& =\left(m y_{1}+a-2\right)\left(m y_{2}+a-2\right)+y_{1} y_{2} \\
& =\left(m^{2}+1\right) y_{1} y_{2}+m(a-2)\left(y_{1}+y_{2}\right)+(a-2)^{2} \\
& =0,
\end{aligned}
$$

Therefore,
$$
\frac{\left(m^{2}+1\right)\left(a^{2}-4\right)}{m^{2}-4}-\frac{2 a m^{2}(a-2)}{m^{2}-4}+(a-2)^{2}=0,
$$

Simplifying, we get $3 a^{2}-16 a+20=0$, solving for $a$ gives $a=2$ (discard) or $a=\frac{10}{3}$. Hence, the line $P Q$ must pass through the fixed point $\left(\frac{10}{3}, 0\right)$.