$\text{ If }{\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}\text{ then the value of }$$x^{100}+y^{100}+z^{100}-\frac{3}{x^{101}+y^{101}+z^{101}}\text{ is }$

$\text{ We have }{\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}.$

$Since -\frac{\mathrm{\pi }}{2}\le s{\mathrm{in}}^{-1}p\le \frac{\pi }{2} \text{∀p∈}\left[-1,1\right]\text{, the above equation is possible when}$

${\sin }^{-1}x=\frac{\pi }{2}\Rightarrow \boxed{x=1}$, ${\sin }^{-1}y=\frac{\pi }{2}\Rightarrow \boxed{y=1} and {\sin }^{-1}z=\frac{\pi }{2}\Rightarrow \boxed{z=1}$

$\therefore x^{100}+y^{100}+z^{100}-\frac{3}{x^{101}+y^{101}+z^{101}}$$=1+1+1-\frac{3}{3}=3-1=2$