If ${\left({\sin }^{-1}x\right)}^2+{\left({\sin }^{-1}y\right)}^2+{\left({\sin }^{-1}z\right)}^2=\frac{3}{4}{\mathrm{\pi }}^2$, find the value of x ² + y ² + z ²

We know that the maximum value of ${\sin }^{-1}x, {\sin }^{-1}y, {\sin }^{-1}z \mathrm{is}\frac{\mathrm{\pi }}{2}$ and minimum value of ${\sin }^{-1}x, {\sin }^{-1}y, {\sin }^{-1}z \mathrm{is}-\frac{\mathrm{\pi }}{2}$
Now,
For maximum value
$$\begin{gathered} \mathrm{LHS}={\left({\sin }^{-1}x\right)}^2+{\left({\sin }^{-1}y\right)}^2+{\left({\sin }^{-1}z\right)}^2 \\ ={\left(\frac{\mathrm{\pi }}{2}\right)}^2+{\left(\frac{\mathrm{\pi }}{2}\right)}^2+{\left(\frac{\mathrm{\pi }}{2}\right)}^2 \\ =\frac{3}{4}{\mathrm{\pi }}^2=\mathrm{RHS} \end{gathered}$$
and For minimum value
$$\begin{gathered} \mathrm{LHS}={\left({\sin }^{-1}x\right)}^2+{\left({\sin }^{-1}y\right)}^2+{\left({\sin }^{-1}z\right)}^2 \\ ={\left(-\frac{\mathrm{\pi }}{2}\right)}^2+{\left(-\frac{\mathrm{\pi }}{2}\right)}^2+{\left(-\frac{\mathrm{\pi }}{2}\right)}^2 \\ =\frac{3}{4}{\mathrm{\pi }}^2=\mathrm{RHS} \end{gathered}$$
Now, For maximum value
$$\begin{gathered} {\sin }^{-1}x=\frac{\mathrm{\pi }}{2}, {\sin }^{-1}y=\frac{\mathrm{\pi }}{2}, {\sin }^{-1}z=\frac{\mathrm{\pi }}{2} \\ \Rightarrow x=\sin \frac{\mathrm{\pi }}{2}, y=\sin \frac{\mathrm{\pi }}{2}, z=\sin \frac{\mathrm{\pi }}{2} \\ \Rightarrow x=1, y=1, z=1 \\ \therefore x^2+y^2+z^2=1+1+1=3 \end{gathered}$$
and for minimum value
$$\begin{gathered} {\sin }^{-1}x=-\frac{\mathrm{\pi }}{2}, {\sin }^{-1}y=-\frac{\mathrm{\pi }}{2}, {\sin }^{-1}z=-\frac{\mathrm{\pi }}{2} \\ \Rightarrow x=\sin \left(-\frac{\mathrm{\pi }}{2}\right), y=\sin \left(-\frac{\mathrm{\pi }}{2}\right), z=\sin \left(-\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow x=-1, y=-1, z=-1 \\ \therefore x^2+y^2+z^2=1+1+1=3 \end{gathered}$$