2. Given $a+b+c=3, a^{2}+b^{2}+c^{2}=3$. Then the value of $a^{2005}+b^{2005}+c^{2005}$ is ( ).
(A) 0
(B) 3
(C) $2^{2005}$
(D) $3 \times 2^{2005}$
2.B.
$$
\begin{array}{l}
\text { Given } 0=3\left(a^{2}+b^{2}+c^{2}\right)-(a+b+c)^{2} \\
=(a-b)^{2}+(b-c)^{2}+(c-a)^{2},
\end{array}
$$

we know $a=b=c=1$.
Thus, $a^{2005}+b^{2005}+c^{2005}=3$.