3. Let $a$, $b$, and $c$ be non-zero complex numbers, and let $\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$. If $\frac{a}{b} = \frac{b}{c} = \frac{c}{a}$, then $\frac{a+b-c}{a-b+c} = (\quad)$.
(A) 1
(B) $\pm \omega$
(C) $1, \omega, \omega^{2}$
(D) $1, -\omega, \omega^{2}$
3. C.

Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=k$. Then
$$
\begin{array}{l}
k^{3}=1 \Rightarrow k=1, \omega, \omega^{2} . \\
\text { Hence } \frac{a+b-c}{a-b+c}=\frac{b k+b-b k^{2}}{b k-b+b k^{2}}=\frac{k+1-k^{2}}{k\left(1-k^{2}+k\right)} \\
=\frac{1}{k}=\left\{\begin{array}{ll}
1, & k=1 ; \\
\omega^{2}, & k=\omega ; \\
\omega, & k=\omega^{2} .
\end{array}\right.
\end{array}
$$