3. If $\frac{x-y}{a}=\frac{y-z}{b}=\frac{z-x}{c}=a b c<0$, then the number of negative numbers among $a, b, c$ is ( ).
(A) 1
(B) 2
(C) 3
(D) Cannot be determined
3. A.

From the given
$$
x-y=a^{2} b c, y-z=a b^{2} c, z-x=a b c^{2} .
$$

Adding them up yields $a b c(a+b+c)=0$. But $a b c \neq 0$, so
$$
a+b+c=0 .
$$

This implies that among $a, b, c$, there is 1 or 2 negative numbers, but since $a b c<0$, there can only be 1 negative number.