1. Let real numbers $x, y, z$ satisfy
$$
x+\frac{y}{z}=y+\frac{z}{x}=z+\frac{x}{y}=2 \text {. }
$$

Then $x y z=$ ( ).
(A) 3 or 7
(B) 1 or 3
(C) -3 or -7
(D) 1 or -7
1. D.
From the problem, we have
$$
\left\{\begin{array}{l}
z x+y=2 z, \\
x y+z=2 x, \\
y z+x=2 y
\end{array}\right.
$$
$$
\Rightarrow x y+y z+z x=x+y+z \triangleq p \text {. }
$$

From $\left\{\begin{array}{l}2-x=\frac{y}{z}, \\ 2-y=\frac{z}{x}, \\ 2-z=\frac{x}{y}\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow(2-x)(2-y)(2-z)=1 \\
\Rightarrow 8-4(x+y+z)+2(x y+y z+z x)-x y z=1 \\
\Rightarrow x y z=7-2 p .
\end{array}
$$

Notice that, the system of equations (1) can be transformed into
$$
\begin{array}{l}
\left\{\begin{array}{l}
x y z=2 y z-y^{2}, \\
x y z=2 z x-z^{2}, \\
x y z=2 x y-x^{2}
\end{array}\right. \\
\Rightarrow 3 x y z=2(x y+y z+z x)-\left(x^{2}+y^{2}+z^{2}\right) \\
\quad=4 p-p^{2} .
\end{array}
$$

Thus, $21-6 p=4 p-p^{2}$
$$
\Rightarrow(p-3)(p-7)=0
$$
$\Rightarrow p=3$ or 7.
Therefore, $x y z=7-2 p=1$ or -7.