Let $\overrightarrow{\mathrm{OA}}=2 \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=6 \overrightarrow{\mathrm{a}}+5 \overrightarrow{\mathrm{~b}}$ and $\overrightarrow{\mathrm{OC}}=3 \overrightarrow{\mathrm{~b}}$, where O is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OC}}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to