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!

Let's rewrite the system as $$\mathbf r \times \mathbf P\mathbf{r} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\x&y&z\\z&x&y\end{vmatrix} = 2\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} \equiv \mathbf f$$
where $\mathbf{r} = (x,y,z)^\top$ and
$$
\mathbf P = \begin{pmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}
$$
So we want to solve
$$
\mathbf{r} \times \mathbf{Pr} = \mathbf{f}.
$$
Let's dot multiply both sides with $\mathbf{r}$
$$
0 = \mathbf{r}^\top \mathbf{f}
$$
also when multiplying by $\mathbf{Pr}$:
$$
0 = \mathbf{r}^\top \mathbf P^\top \mathbf{f}
$$
That's a pair of linear equations in $x,y,z$. Since the equations are homogeneous, the solution will be of the form $\mathbf r = \alpha \mathbf{h}, \alpha \in \mathbb{R}$.
$$
2x + 3y + z = 0\\
2z + 3x + y = 0
$$
Row reducing the system we get
$$
\mathbf h = \begin{pmatrix}
-5\\1\\7
\end{pmatrix}.
$$
Since $\mathbf h \times \mathbf {Ph} = 18 \mathbf f$ we deduce that
$$
\alpha^2 = \frac{1}{18}\qquad\alpha = \pm\frac{1}{3\sqrt{2}}.
$$
Finally
$$
x = \mp \frac{5}{3\sqrt{2}}\\
y = \pm \frac{1}{3\sqrt{2}}\\
z = \pm \frac{7}{3\sqrt{2}}
$$
