11. 设点OOO在ΔABC\Delta ABCΔABC内部,且有OA→+2OB→+3OC→=0→\overrightarrow{OA}+2\overrightarrow{OB}+3\overrightarrow{OC}=\overrightarrow{0}OA+2OB+3OC=0,点DDD是边BCBCBC的中点,设ΔADC\Delta ADCΔADC与ΔAOC\Delta AOCΔAOC的面积分别为S1S_{1}S1、S2S_{2}S2,则S1:S2=(S_{1}:S_{2}=(S1:S2=( )))