Given,
\(\begin{aligned}
z & =x+i y=\frac{(3+2 i)(4-7 i)(12+13 i)}{(13-12 i)(2-3 i)(11+3 i)} \\
|z| & =\frac{|3+2 i| \cdot|4-7 i| \cdot|12+13 i|}{|13-12 i| \cdot|2-3 i| \cdot|11+3 i|} \\
& =\frac{\left(\sqrt{3^2+2^2}\right)\left(\sqrt{4^2+7^2}\right)\left(\sqrt{\left(12^2+13^2\right.}\right)}{\left(\sqrt{13^2+12^2}\right) \cdot\left(\sqrt{2^2+3^2}\right)\left(\sqrt{11^2+3^2}\right)}
\end{aligned}\)
\(\begin{aligned}
= & \frac{\sqrt{4^2+7^2}}{\sqrt{11^2+3^2}}=\frac{\sqrt{65}}{\sqrt{130}}=\frac{1}{\sqrt{2}} \\
\Rightarrow \quad \sqrt{x^2+y^2} & =\frac{1}{\sqrt{2}} \text { or } x^2+y^2=\frac{1}{2}
\end{aligned}\)