If the general solution of the equation \(\frac{\tan 3 x-1}{\tan 3 x+1}=\sqrt{3}\) is \(x=\frac{\mathrm{n} \pi}{\mathrm{p}}+\frac{7 \pi}{\mathrm{q}}, \mathrm{n}, \mathrm{p}, \mathrm{q}, \in \mathrm{Z}\), then \(\frac{p}{q}\) is

\( \frac{\tan 3 x-1}{\tan 3 x+1}=\sqrt{3} \)
\( \therefore \frac{\tan 3 x-\tan \frac{\pi}{4}}{1+(\tan 3 x)\left(\tan \frac{\pi}{4}\right)}=\sqrt{3} \)
\( \therefore \tan \left(3 x-\frac{\pi}{4}\right)=\tan \left(\frac{\pi}{3}\right) \)
\( \therefore 3 x-\frac{\pi}{4}=n \pi+\frac{\pi}{3}\)
\(\therefore 3 x=\mathrm{n} \pi+\frac{\pi}{3}+\frac{\pi}{4} \)
\(\therefore 3 x=\mathrm{n} \pi+\frac{7 \pi}{12} \)
\(\therefore x=\frac{\mathrm{n} \pi}{3}+\frac{7 \pi}{36}\)
Comparing with \(\frac{\mathrm{n} \pi}{\mathrm{p}}+\frac{7 \pi}{\mathrm{q}}\), we get
\(\mathrm{p}=3, \mathrm{q}=36\)
\(\therefore \frac{\mathrm{p}}{\mathrm{q}}=\frac{3}{36}=\frac{1}{12}\)