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

$$\vec{OA} = 2\vec{a}$$, $$\vec{OB} = 6\vec{a} + 5\vec{b}$$, $$\vec{OC} = 3\vec{b}$$.

Area of parallelogram with sides $$\vec{OA}$$ and $$\vec{OC}$$ = $$|2\vec{a} \times 3\vec{b}| = 6|\vec{a} \times \vec{b}| = 15$$.

So $$|\vec{a} \times \vec{b}| = \frac{15}{6} = \frac{5}{2}$$.

Area of quadrilateral OABC = Area of triangle OAB + Area of triangle OBC.

Area of $$\triangle OAB = \frac{1}{2}|\vec{OA} \times \vec{OB}| = \frac{1}{2}|2\vec{a} \times (6\vec{a}+5\vec{b})| = \frac{1}{2}|10(\vec{a}\times\vec{b})| = 5|\vec{a}\times\vec{b}| = \frac{25}{2}$$.

Area of $$\triangle OBC = \frac{1}{2}|\vec{OB} \times \vec{OC}| = \frac{1}{2}|(6\vec{a}+5\vec{b}) \times 3\vec{b}| = \frac{1}{2}|18(\vec{a}\times\vec{b})| = 9|\vec{a}\times\vec{b}| = \frac{45}{2}$$.

Total area = $$\frac{25}{2} + \frac{45}{2} = \frac{70}{2} = 35$$.

The correct answer is Option 4: 35.