If  ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\frac{\pi }{2}$and  ${(x-y)}^2+{(y-z)}^2+{(z-x)}^2=0$ then  $x^2+y^2+z^2$is

$$\begin{gathered} {\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z=\frac{\mathrm{\pi }}{2} \\ \Rightarrow {\tan }^{-1}\left(\frac{x+y+z-xyz}{1-\left(xy+yz+zx\right)}\right)=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \frac{x+y+z-xyz}{1-\left(xy+yz+zx\right)}=\tan \frac{\mathrm{\pi }}{2}=not defined \\ \Rightarrow 1-\left(xy+yz+zx\right)=0 \\ \Rightarrow xy+yz+zx=1 \\ Now, {\left(x-y\right)}^2+{\left(y-z\right)}^2+{\left(z-x\right)}^2=0 \\ \Rightarrow 2\left(x^2+y^2+z^2-\left(xy+yz+zx\right)\right)=0 \\ \Rightarrow x^2+y^2+z^2-\left(xy+yz+zx\right)=0 \\ \Rightarrow x^2+y^2+z^2=1 \end{gathered}$$