Let $\overrightarrow { \mathrm { OA } } = \overrightarrow { \mathrm { a } } , \overrightarrow { \mathrm { OB } } = 12 \overrightarrow { \mathrm { a } } + 4 \overrightarrow { \mathrm {~b} }$ and $\overrightarrow { \mathrm { OC } } = \overrightarrow { \mathrm { b } }$, where O is the origin. If $S$ is the parallelogram with adjacent sides OA and OC, then $\frac { \text { area of the quadrilateral } \mathrm { OABC } } { \text { area of } \mathrm { S } }$ is equal to (1) 6 (2) 10 (3) 7 (4) 8
Let $\overrightarrow { \mathrm { OA } } = \overrightarrow { \mathrm { a } } , \overrightarrow { \mathrm { OB } } = 12 \overrightarrow { \mathrm { a } } + 4 \overrightarrow { \mathrm {~b} }$ and $\overrightarrow { \mathrm { OC } } = \overrightarrow { \mathrm { b } }$, where O is the origin. If $S$ is the parallelogram with adjacent sides OA and OC, then $\frac { \text { area of the quadrilateral } \mathrm { OABC } } { \text { area of } \mathrm { S } }$ is equal to\\
(1) 6\\
(2) 10\\
(3) 7\\
(4) 8