1. Given that $a, b, c$ are all positive numbers and none of them equals 1. If real numbers $x, y, z$ satisfy $a^{x}=b^{y}=c^{z}, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ $=0$, then the value of $a b c$ is $(\quad)$.
(A) $\frac{1}{2}$
(B) 1
(C) 2
(D) 4
-、1.B.
Let $a^{x}=b^{y}=c^{z}=t(t>0$, and $t \neq 1)$, then
$$
\frac{1}{x}=\log _{t} a, \frac{1}{y}=\log _{t} b, \frac{1}{z}=\log _{t} c \text {. }
$$

Since $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\log _{6} a+\log _{6} b+\log _{2} c$ $=\log _{1}(a b c)=0$,
therefore, $a b c=1$.