If the angle made by the tangent at the point $$(x_0, y_0)$$ on the curve $$x = 12(t + \sin t \cos t), y = 12(1 + \sin t)^2, 0 < t < \frac{\pi}{2}$$, with the positive $$x$$-axis is $$\frac{\pi}{3}$$, then $$y_0$$ is equal to

We are given the parametric curve $$x = 12(t + \sin t \cos t)$$ and $$y = 12(1 + \sin t)^2$$ where $$0 < t < \frac{\pi}{2}$$.

The slope of the tangent is $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$.

First, compute $$\frac{dx}{dt}$$:

$$\frac{dx}{dt} = 12(1 + \cos^2 t - \sin^2 t) = 12(1 + \cos 2t) = 24\cos^2 t$$

Next, compute $$\frac{dy}{dt}$$:

$$\frac{dy}{dt} = 12 \cdot 2(1 + \sin t) \cdot \cos t = 24(1 + \sin t)\cos t$$

Therefore:

$$\frac{dy}{dx} = \frac{24(1 + \sin t)\cos t}{24\cos^2 t} = \frac{1 + \sin t}{\cos t}$$

Since the tangent makes angle $$\frac{\pi}{3}$$ with the positive $$x$$-axis:

$$\frac{1 + \sin t}{\cos t} = \tan\frac{\pi}{3} = \sqrt{3}$$

So $$1 + \sin t = \sqrt{3}\cos t$$.

Rearranging: $$\sqrt{3}\cos t - \sin t = 1$$

This can be written as $$2\sin\left(\frac{\pi}{3} - t\right) = 1$$, giving $$\sin\left(\frac{\pi}{3} - t\right) = \frac{1}{2}$$.

Since $$0 < t < \frac{\pi}{2}$$, we get $$\frac{\pi}{3} - t = \frac{\pi}{6}$$, so $$t = \frac{\pi}{6}$$.

Now compute $$y_0$$:

$$y_0 = 12\left(1 + \sin\frac{\pi}{6}\right)^2 = 12\left(1 + \frac{1}{2}\right)^2 = 12 \cdot \frac{9}{4} = 27$$

The answer is Option C: $$27$$.