Correct Answer: $6 \sqrt{3}$

Solution :

O is a point in the interior of $\triangle \mathrm{ABC}$
OA = 12 cm
OC = 9 cm
$\angle A O B=\angle B O C=\angle C O A$
$\angle A B C=60^{\circ}$
$\angle A O B=\angle B O C=\angle C O A=\frac{360^{\circ}}{3} = 120^{\circ}$
Let $\angle OBA = \theta$
In $\triangle AOB$,
$\angle OBA + \angle BAO + \angle AOB = 180^{\circ}$
⇒ $\theta + \angle BAO + 120^{\circ} = 180^{\circ}$
⇒ $\angle BAO = 60^{\circ} - \theta$
In $\triangle BOC$,
$\angle OBC = \angle ABC - \angle OBA$
⇒ $ \angle OBC = 60^{\circ} - \theta$
⇒ $\angle OBC + \angle BOC + \angle OCB = 180^{\circ}$
⇒ $60^{\circ} - \theta + 120^{\circ} + \angle OCB = 180^{\circ}$
⇒ $\angle OCB = 180^{\circ}-120^{\circ}-60^{\circ}-\theta$
⇒ $\angle OCB = \theta$
In $\triangle AOB$ and $\triangle BOC$,
$\angle A = \angle B$
$\angle O = \angle O$
$\angle B = \angle C$
By AAA similarity, $\triangle AOB$ ~ $\triangle BOC$
⇒ $\frac{OA}{OB} = \frac{OB}{OC}$
⇒ $\frac{12}{OB} = \frac{OB}{9}$
⇒ $OB^2 = 12 × 9$
⇒ $OB = \sqrt{108} = 6\sqrt3$
Hence, the correct answer is $6\sqrt3$.