B. Let non-zero real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+2 b+3 c=0, \\
2 a+3 b+4 c=0 .
\end{array}\right. \\
\text { Then } \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}=(\quad \text {. }
\end{array}
$$
(A) $-\frac{1}{2}$
(B) 0
(C) $\frac{1}{2}$
(D) 1
B. A.

From the given we have
$$
\begin{array}{l}
a+b+c=(2 a+3 b+4 c)-(a+2 b+3 c)=0 \\
\Rightarrow(a+b+c)^{2}=0 \\
\Rightarrow a b+b c+c a=-\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right) \\
\Rightarrow \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}=-\frac{1}{2} .
\end{array}
$$