$$
\begin{aligned}
& O P: x-3 y-5=0 \\
& O Q: x+5 y-9=0 \\
& \therefore \quad O\left(\frac{13}{2}, \frac{1}{2}\right)
\end{aligned}
$$

Since $P$ is mid point of $A B$. Hence $P\left(\frac{1+x_1}{2}, \frac{2+y_1}{2}\right)$
$\because \quad P$ lies on $O P$ and $O P \perp A B$.
$$
\begin{aligned}
& \therefore \quad \frac{1+x_1}{2}-3\left(\frac{2+y_1}{2}\right)-5=0 \\
& \Rightarrow 1+x_1-3\left(2+y_1\right)-10=0
\end{aligned}
$$
$\Rightarrow x_1-3 y_1=15$ ...(1)
and $m_{O P} \times m_{A B}=-1$
$$
\Rightarrow \frac{1}{3} \times \frac{y_1-2}{x_1-1}=-1 \Rightarrow-3\left(x_1-1\right)=y_1-2
$$
$\Rightarrow 3 x_1+y_1=5$ ...(2)
From (1) and (2) $\left(x_1, y_1\right) \equiv(3,-4)$
Similarly $Q$ is mid point of $B C$ and $O Q \perp B C$,
$$
Q\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) \equiv Q\left(\frac{3+x_2}{2}, \frac{-4+y_2}{2}\right)
$$
$Q$ lies on $B C$,
$$
\begin{aligned}
& \frac{3+x_2}{2}+\frac{5\left(-4+y_2\right)}{2}-9=0 \\
& \Rightarrow x_2+3-20+5 y_2-18=0
\end{aligned}
$$
$\Rightarrow x_2+5 y_2=35$ ...(3)
and $M_{O Q} \times M_{B C}=-1$
$$
\Rightarrow \frac{-1}{5} \times \frac{y_2+4}{x_2-3}=-1 \Rightarrow 5 x_2-15=y_2+4
$$
$\Rightarrow 5 x_2-y_2=19$ ...(4)
$\begin{aligned} & \text { From (3) and (4), }\left(x_2, y_2\right) \equiv(5,6) \\ & \begin{aligned} \therefore \quad C A & =\sqrt{\left(x_2-1\right)^2+\left(y_2-6\right)^2} \\ & =\sqrt{(5-1)^2+(6-2)^2}=\sqrt{4^2+4^2}=\sqrt{32} \\ \therefore \quad A C & =4 \sqrt{2} .\end{aligned}\end{aligned}$