In the quadrilateral $OACB$, Area of $OACB=p=$Area of $\Delta OAC+$ Area of $\Delta OBC$
$\Rightarrow\:p=\large\frac{1}{2}$$(\overrightarrow {OA}\times \overrightarrow {OC})+\large\frac{1}{2}$$(\overrightarrow {OC}\times \overrightarrow {OB})$
$=\large\frac{1}{2}$$(-\overrightarrow {OC} \times\overrightarrow {OA}+\overrightarrow OC\times\overrightarrow {OB})$
$=\large\frac{1}{2}$$\overrightarrow {OC}\times(\overrightarrow {OB}-\overrightarrow {OA})$
$=\large\frac{1}{2}$$(\overrightarrow {OC}\times \overrightarrow {AB})$
$=\large\frac{1}{2}$$(\overrightarrow b\times(6\overrightarrow a+2\overrightarrow b)$)
$=\large\frac{1}{2}$.$6(\overrightarrow b\times\overrightarrow a)$
Area of parallelogram with sides $\overrightarrow {OA}$ and $\overrightarrow {OC}$
$=q=\large\frac{1}{2}$$(\overrightarrow b\times\overrightarrow b)$
$\therefore\:p=6q$
$\therefore\:x=6$