Lat $(a,b,c)\in S^2$, where $S^2:=\{(x_1,x_2,x_3)\in \mathbb R^3: x_1^2+x_2^2+x_3^2=1 \}$. How to solve the following system of equations: $$ x^2+y^2-z^2-t^2=a,\\ 2(yz+xt)=b,\\ 2(yt-xz)=c,\\ x^2+y^2+z^2+t^2=1 $$ with unknown $x,y,z,t \in \mathbb R$.
Edit. The queston was edited. I added the last equation $$ x^2+y^2+z^2+t^2=1. $$
Assume first that $a \neq -1$.
By the 1. and 4. equation we have $$ x^2+y^2=\frac{a+1}{2}. $$ Then $$ x=\sqrt{\frac{a+1}{2}}\cos \phi,\\ y=\sqrt{\frac{a+1}{2}} \sin \phi, $$ where $\phi$ is an arbitrary numer such that $\phi \in [0,2\pi)$. By the 2. and 3. equations we have $$ 2\sqrt{\frac{a+1}{2}} (\sin \phi) z+2\sqrt{\frac{a+1}{2}} (\cos \phi) t=b,\\ -2\sqrt{\frac{a+1}{2}} (\cos \phi) z +2\sqrt{\frac{a+1}{2}} (\sin \phi) t=c. $$ This gives $$ z=\frac{1}{\sqrt {2(a+1)} }b \sin \phi-\frac{1}{\sqrt {2(a+1)} }c \cos \phi, \\ t=\frac{1}{\sqrt {2(a+1)} }b \cos \phi +\frac{1}{\sqrt {2(a+1)} } \sin \phi. $$
Assume now that $a=-1$. By the same method as above we obtain: $$ x=0,y=0, z=\cos \phi, t=\sin \phi, $$ where $\phi$ is an arbitrary numer such that $\phi \in [0,2\pi)$.