Let $$E:\frac{x^2}{36}+\frac{y^2}{16}=1$$ and $$C$$ be its auxiliary circle. $$AB$$ is a chord of $$E$$. $$A', B'$$ are corresponding points of $$A, B$$ respectively on $$C$$. If $$\angle A'OB'=\frac{\pi}{3}$$ and the slope of $$AB$$ is $$\frac{1}{\sqrt{3}}$$, then what is the value of $$AB^2$$?

The parametric coordinates of points $$A$$ and $$B$$ on the ellipse $$\frac{x^2}{36} + \frac{y^2}{16} = 1$$ are $$(6 \cos \theta_1, 4 \sin \theta_1)$$ and $$(6 \cos \theta_2, 4 \sin \theta_2)$$ respectively.

The corresponding points on the auxiliary circle $$x^2 + y^2 = 36$$ are $$A'(6 \cos \theta_1, 6 \sin \theta_1)$$ and $$B'(6 \cos \theta_2, 6 \sin \theta_2)$$.

Given that the angle $$\angle A'OB' = \frac{\pi}{3}$$, the difference between their eccentric angles is:

$$\vert{}\theta_1 - \theta_2\vert{} = \frac{\pi}{3}$$

The slope $$m$$ of the chord $$AB$$ is expressed as:

$$m = \frac{4 \sin \theta_2 - 4 \sin \theta_1}{6 \cos \theta_2 - 6 \cos \theta_1} = -\frac{2}{3} \cot\left(\frac{\theta_1 + \theta_2}{2}\right)$$

Equating this to the given slope $$\frac{1}{\sqrt{3}}$$:

$$-\frac{2}{3} \cot\left(\frac{\theta_1 + \theta_2}{2}\right) = \frac{1}{\sqrt{3}} \implies \cot\left(\frac{\theta_1 + \theta_2}{2}\right) = -\frac{\sqrt{3}}{2}$$

Letting $$\phi = \frac{\theta_1 + \theta_2}{2}$$, we get $$\tan \phi = -\frac{2}{\sqrt{3}}$$.

Using trigonometric identities, we find $$\cos^2 \phi$$ and $$\sin^2 \phi$$:

$$\cos^2 \phi = \frac{1}{1 + \tan^2 \phi} = \frac{1}{1 + \frac{4}{3}} = \frac{3}{7}$$

$$\sin^2 \phi = 1 - \frac{3}{7} = \frac{4}{7}$$

The square of the length of chord $$AB$$ is expanded using trigonometric differences for $$\theta_2 - \theta_1 = \frac{\pi}{3}$$:

$$AB^2 = 36(\cos \theta_2 - \cos \theta_1)^2 + 16(\sin \theta_2 - \sin \theta_1)^2$$

$$AB^2 = 36 \sin^2\phi + 16 \cos^2\phi$$

Substituting the values of $$\sin^2 \phi$$ and $$\cos^2 \phi$$:

$$AB^2 = 36\left(\frac{4}{7}\right) + 16\left(\frac{3}{7}\right) = \frac{144 + 48}{7} = \frac{192}{7}$$

The correct option is C.