3. Solve the system $\left\{\begin{array}{l}x^{2}=2 \sqrt{y^{2}+1} ; \\ y^{2}=2 \sqrt{z^{2}-1}-2 ; \\ z^{2}=4 \sqrt{x^{2}+2}-6 .\end{array}\right.$
Answer: $(\sqrt{2}, 0, \sqrt{2}),(\sqrt{2}, 0,-\sqrt{2}),(-\sqrt{2}, 0, \sqrt{2}),(-\sqrt{2}, 0,-\sqrt{2})$.

Solution: Let $a=\sqrt{x^{2}+2}, b=\sqrt{y^{2}+1}, c=\sqrt{z^{2}-1}$. We get the system: $\left\{\begin{array}{l}a^{2}-2=2 b \\ b^{2}-1=2 c-2 \\ c^{2}+1=4 a-6\end{array}\right.$

Add all the equations and move to the left side: $a^{2}+b^{2}+c^{2}-4 a-2 b-2 c+6=0$. By completing the squares, we get $(a-2)^{2}+(b-1)^{2}+(c-1)^{2}=0$, from which $a=2$, $b=1, c=1$. Substituting back, we get $x= \pm \sqrt{2}, y=0, z= \pm \sqrt{2}$.