jmc

To find the area of $△ABC$ in terms of $p$, we find the area of $ABOQ$ and subtract out the areas of $△ACQ$ and $△BCO.$

Both $\overline{QA}$ and $\overline{OB}$ are horizontal, so $\overline{QA}$ is parallel to $\overline{OB}$. Thus, $ABOQ$ is a trapezoid with bases $\overline{AQ}$ and $\overline{OB}.$ Since $\overline{OQ}$ is vertical, its length is the height of the trapezoid, so the area of $ABOQ$ is $$\frac{1}{2}\cdot QO\cdot (QA+OB)=\frac{1}{2}\cdot 12\cdot (2+12)=84.$$Since $△ACQ$ has a right angle at $Q,$ its area is $$\frac{1}{2}\cdot QA\cdot QC=\frac{1}{2}\cdot (2-0)\cdot (12-p)=12-p.$$Since $△COB$ has a right angle at $O,$ its area is $$\frac{1}{2}\cdot OB\cdot CO=\frac{1}{2}\cdot (12-0)\cdot (p-0)=6p.$$Thus, the area of $△ABC$ is $$84-6p-(12-p)=72-5p.$$Then $72-5p=27$ or $5p=45,$ so $p=\boxed{9}.$