点 $\mathbf{D}$ 为 $\triangle A B C$ 内一点, 且 $\overrightarrow{D A}+4 \overrightarrow{D B}+7 \overrightarrow{D C}=\overrightarrow{0}$, 则 $\frac{S_{\triangle B C D}}{S_{\triangle A B C}}=(\quad)$
<ImageHere>分别延长 $D B, D C$ 至 $B_{1}, C_{1}$, 使得 $D B_{1}=3 D B, D C_{1}=7 D C$, 则 $\overrightarrow{D A}+\overrightarrow{D B_{1}}+\overrightarrow{D C_{1}}=\overrightarrow{0}$, 则 $S_{\triangle D A B_{1}}=S_{\triangle D A G_{1}}=S_{\triangle D B_{1} G_{1}}=S$, $S_{\triangle D A B}=\frac{1}{4} s, S_{\triangle D A C}=\frac{1}{7} s, S_{\triangle D B C}=\frac{1}{28} s, S_{\triangle A B C}=\frac{1}{4} s+\frac{1}{7} s+\frac{1}{28} s=\frac{12}{28} s, \frac{S_{\triangle B C D}}{S_{\triangle A B C}}=\frac{\frac{1}{28} s}{\frac{12}{28} s}=\frac{1}{12}$, 故选 D.