The minimum value of $\sqrt{(3\sin x-4\cos x-10)(3\sin x+4\cos x-10)}$ is

$f\left(x\right)=9{\sin }^{2}x-16{\cos }^{2}x$

$-10(3\sin x-4\cos x)-10(3\sin x+4\cos x)+100$

$=25{\sin }^{2}x-60\sin x+84={(5\sin x-6)}^2+48$

$\therefore$ the minimum value of $f\left(x\right)$ occurs when $\sin x=1\therefore$ minimum value of $\sqrt{f\left(x\right)}$ is 7.