If $\sin \frac{23\pi }{24}=\sqrt{\frac{2\sqrt{p}-\sqrt{q}-1}{4\sqrt{r}}}$ ,   then $p^2+q^2-r^2=$

$\cos \left(\frac{\pi }{12}\right)=\cos 15°=\frac{\sqrt{3}+1}{2\sqrt{2}}$

$1-\cos \theta =2{\sin }^2\left(\frac{\theta }{2}\right)$

$\Rightarrow 2{\sin }^2\left(\frac{\pi }{24}\right)=1-\cos \left(\frac{\pi }{12}\right)$

$\Rightarrow {\sin }^2\left(\frac{\pi }{24}\right)=\frac{1-\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}{2}$

$\Rightarrow \sin \left(\frac{\pi }{24}\right)=\sqrt{\frac{2\sqrt{2}-\sqrt{3}-1}{4\sqrt{2}}}$

$p=2,q=3,r=2$

$\therefore p^2+q^2-r^2=2^2+3^2-2^2=9$