3. If real numbers $a, b, c$ satisfy the condition $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ $=\frac{1}{a+b+c}$, then among $a, b, c$ (   ).
(A) there must be two numbers that are equal
(B) there must be two numbers that are opposites
(C) there must be two numbers that are reciprocals
(D) no two numbers are equal
3. B.

From the given, we have $(a+b+c)(bc+ca+ab)=abc$.
Expanding and rearranging, we get
$$
\begin{array}{l}
a^{2} c+a^{2} b+b^{2} c+b^{2} a+c^{2} b+c^{2} a+2 a b c=0 \\
\Rightarrow a c(a+b)+a b(a+b)+b c(a+b)+c^{2}(a+b)=0 \\
\Rightarrow a(a+b)(b+c)+c(a+b)(b+c)=0 \\
\Rightarrow(a+b)(b+c)(c+a)=0 .
\end{array}
$$

Thus, at least one of $a+b$, $b+c$, and $c+a$ is 0.
Therefore, among $a$, $b$, and $c$, there must be two numbers that are opposites of each other.