First, calculate the length of OC and AC using trigonometry. Since \(\angle AOB = \frac{1}{3} \pi\), we have:

\(OC = 12 \cos\left(\frac{1}{3} \pi\right) = 12 \times \frac{1}{2} = 6\)

\(AC = 12 \sin\left(\frac{1}{3} \pi\right) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\)

The area of the sector AOB is given by:

\(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 12^2 \times \frac{1}{3} \pi = 24\pi\)

The area of rectangle OCDB is:

\(12 \times 6\sqrt{3} = 72\sqrt{3}\)

The area of triangle OAC is:

\(\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\)

The area of the shaded region is the area of the rectangle minus the area of the sector and the triangle:

\(72\sqrt{3} - (24\pi + 18\sqrt{3}) = 54\sqrt{3} - 24\pi\)

Thus, the values of a and b are 54 and 24, respectively.