Question:

Step 1: Determine the magnitude of the cross product of vectors $\vec{a}$ and $\vec{b}$. The area of a parallelogram with adjacent sides $\vec{u}$ and $\vec{v}$ is given by $|\vec{u} \times \vec{v}|$. Given that the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is 15 sq. units. We have $\overrightarrow{OA} = 2\vec{a}$ and $\overrightarrow{OC} = 3\vec{b}$. Therefore, the area of the parallelogram is: $$ \text{Area} = |\overrightarrow{OA} \times \overrightarrow{OC}| = |(2\vec{a}) \times (3\vec{b})| $$ $$ = |2 \cdot 3 (\vec{a} \times \vec{b})| = |6 (\vec{a} \times \vec{b})| = 6 |\vec{a} \times \vec{b}| $$ Given that this area is 15 sq. units, we have: $$ 6 |\vec{a} \times \vec{b}| = 15 $$ $$ |\vec{a} \times \vec{b}| = \frac{15}{6} = \frac{5}{2} $$ Step 2: Decompose the quadrilateral $OABC$ into triangles. The quadrilateral $OABC$ can be divided into two triangles by connecting points $O$ and $B$. These triangles are $\triangle OAB$ and $\triangle OBC$. The total area of the quadrilateral $OABC$ will be the sum of the areas of these two triangles: $$ \text{Area}(OABC) = \text{Area}(\triangle OAB) + \text{Area}(\triangle OBC) $$ Step 3: Calculate the area of $\triangle OAB$. The area of a triangle with adjacent sides $\vec{u}$ and $\vec{v}$ is given by $\frac{1}{2}|\vec{u} \times \vec{v}|$. For $\triangle OAB$, the adjacent sides originating from $O$ are $\overrightarrow{OA}$ and $\overrightarrow{OB}$. We have $\overrightarrow{OA} = 2\vec{a}$ and $\overrightarrow{OB} = 6\vec{a} + 5\vec{b}$. $$ \text{Area}(\triangle OAB) = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}| = \frac{1}{2} |(2\vec{a}) \times (6\vec{a} + 5\vec{b})| $$ $$ = \frac{1}{2} |(2\vec{a} \times 6\vec{a}) + (2\vec{a} \times 5\vec{b})| $$ Using the property $\vec{x} \times \vec{x} = \vec{0}$ and the linearity of the cross product: $$ = \frac{1}{2} |12(\vec{a} \times \vec{a}) + 10(\vec{a} \times \vec{b})| $$ $$ = \frac{1}{2} |\vec{0} + 10(\vec{a} \times \vec{b})| = \frac{1}{2} |10(\vec{a} \times \vec{b})| $$ $$ = \frac{1}{2} \cdot 10 |\vec{a} \times \vec{b}| = 5 |\vec{a} \times \vec{b}| $$ Substitute the value of $|\vec{a} \times \vec{b}| = \frac{5}{2}$ from Step 1: $$ \text{Area}(\triangle OAB) = 5 \cdot \frac{5}{2} = \frac{25}{2} $$ Step 4: Calculate the area of $\triangle OBC$. For $\triangle OBC$, the adjacent sides originating from $O$ are $\overrightarrow{OB}$ and $\overrightarrow{OC}$. We have $\overrightarrow{OB} = 6\vec{a} + 5\vec{b}$ and $\overrightarrow{OC} = 3\vec{b}$. $$ \text{Area}(\triangle OBC) = \frac{1}{2} |\overrightarrow{OB} \times \overrightarrow{OC}| = \frac{1}{2} |(6\vec{a} + 5\vec{b}) \times (3\vec{b})| $$ $$ = \frac{1}{2} |(6\vec{a} \times 3\vec{b}) + (5\vec{b} \times 3\vec{b})| $$ Using the property $\vec{x} \times \vec{x} = \vec{0}$: $$ = \frac{1}{2} |18(\vec{a} \times \vec{b}) + 15(\vec{b} \times \vec{b})| $$ $$ = \frac{1}{2} |18(\vec{a} \times \vec{b}) + \vec{0}| = \frac{1}{2} |18(\vec{a} \times \vec{b})| $$ $$ = \frac{1}{2} \cdot 18 |\vec{a} \times \vec{b}| = 9 |\vec{a} \times \vec{b}| $$ Substitute the value of $|\vec{a} \times \vec{b}| = \frac{5}{2}$ from Step 1: $$ \text{Area}(\triangle OBC) = 9 \cdot \frac{5}{2} = \frac{45}{2} $$ Step 5: Calculate the total area of the quadrilateral $OABC$. Using the result from Step 2, sum the areas of $\triangle OAB$ and $\triangle OBC$: $$ \text{Area}(OABC) = \text{Area}(\triangle OAB) + \text{Area}(\triangle OBC) $$ $$ = \frac{25}{2} + \frac{45}{2} = \frac{25+45}{2} = \frac{70}{2} = 35 $$ Step 6: State the final answer. The area of the quadrilateral $OABC$ is 35 sq. units. The final answer is $\boxed{35}$.