8. If $a+b+c=0, \frac{1}{a+1}+\frac{1}{b+2}+$ $\frac{1}{c+3}=0$, then, $(a+1)^{2}+(b+2)^{2}+(c+3)^{2}$ is equal to (.
(A) 36
(B) 16
(C) 14
(D) 3
8. A.

From the problem, we have
$$
\begin{array}{l}
(a+1)(b+2)+(b+2)(c+3)+(c+3)(a+1)=0 \text {. } \\
\text { Then }(a+1)^{2}+(b+2)^{2}+(c+3)^{2} \\
=(a+b+c+6)^{2}-2[(a+1)(b+2)+ \\
(b+2)(c+3)+(c+3)(a+1)]=36 .
\end{array}
$$