Solving a system of non-linear equations Let
$$(\star)\begin{cases}
 \begin{vmatrix}
 x&y\\
 z&x\\
 \end{vmatrix}=1,   \\ 
 \begin{vmatrix}
 y&z\\
 x&y\\
 \end{vmatrix}=2,  \\ 
 \begin{vmatrix}
 z&x\\
 y&z\\
 \end{vmatrix}=3.  
\end{cases}$$ 
Solving the above system of three non-linear equations with three unknowns.

I have a try. 
Let$$A=\begin{bmatrix}
 1&  1/2& -1/2\\ 
 1/2&  1& -1/2\\ 
 -1/2& -1/2& -1 
\end{bmatrix}$$
We have $$(x,y,z)A\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=0.$$
There must be a orthogonal matrix $T$,such that $T^{-1}A T=diag \begin{Bmatrix}
\frac{1}{2},\frac{\sqrt{33}+1}{4},-\frac{\sqrt{33}-1}{4}
\end{Bmatrix}.$
$$\begin{pmatrix}
x\\ 
y\\ 
z
\end{pmatrix}=T\begin{pmatrix}
x^{'}\\ 
y^{'}\\ 
z^{'}
\end{pmatrix}\Longrightarrow\frac{1}{2} {x'}^{2}+\frac{\sqrt{33}+1}{4} {y'}^{2}-\frac{\sqrt{33}-1}{4}{z'}^{2}=0.$$
But even if we find a $\begin{pmatrix}
x_0^{'}\\ 
y_0^{'}\\ 
z_0^{'}
\end{pmatrix} $ satisfying $\frac{1}{2} {x_0'}^{2}+\frac{\sqrt{33}+1}{4} {y_0'}^{2}-\frac{\sqrt{33}-1}{4}{z_0'}^{2}=0,\begin{pmatrix}
x_0\\ 
y_0\\ 
z_0
\end{pmatrix}=T\begin{pmatrix}
x_0^{'}\\ 
y_0^{'}\\ 
z_0^{'}
\end{pmatrix}$ may not be the solution of $(\star)$

If you have some  good ideas,please give me some hints. Any help would be appreciated!

The resultant of $x^2-yz-1$ and $y^2-xz-2$ with respect to $z$ is $x^3-y^3-x+2y$.  The resultant of $x^2 - yz - 1$ and $z^2 - xy - 3$ with respect to $z$ is $x^4-x y^3-2 x^2-3 y^2+1$.  The resultant of $x^3-y^3-x+2y$ and 
$x^4-x y^3-2 x^2-3 y^2+1$ with respect to $x$ is $-y^4(18 y^2-1)$.  So either
$y = 0$ or $y = \pm 1/\sqrt{18}$.
With $y=0$ we get $x^2 - 1 = 0$, $-xz-2 = 0$ and $z^2 - 3 = 0$, which clearly will not work.
With $y = \pm 1/\sqrt{18} = \sqrt{2}/6$ we do get 
solutions: $x = \mp 5 \sqrt{2}/6$, $z = \pm 7 \sqrt{2}/6$.
