Consider three vectors $$\vec{a}, \vec{b}, \vec{c}$$. Let $$|\vec{a}| = 2, |\vec{b}| = 3$$ and $$\vec{a} = \vec{b} \times \vec{c}$$. If $$\alpha \in \left[0, \frac{\pi}{3}\right]$$ is the angle between the vectors $$\vec{b}$$ and $$\vec{c}$$, then the minimum value of $$27|\vec{c} - \vec{a}|^2$$ is equal to:

Given $$|\vec{a}| = 2$$, $$|\vec{b}| = 3$$, $$\vec{a} = \vec{b} \times \vec{c}$$, and $$\alpha \in [0, \pi/3]$$ is the angle between $$\vec{b}$$ and $$\vec{c}$$.

Since $$\vec{a} = \vec{b} \times \vec{c}$$, we have $$|\vec{a}| = |\vec{b}||\vec{c}|\sin\alpha$$.

$$ 2 = 3|\vec{c}|\sin\alpha \Rightarrow |\vec{c}| = \frac{2}{3\sin\alpha} $$

Also, $$\vec{a} \perp \vec{b}$$ and $$\vec{a} \perp \vec{c}$$.

Compute $$|\vec{c} - \vec{a}|^2$$.

$$ |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 - 2\vec{a} \cdot \vec{c} + |\vec{a}|^2 $$

Since $$\vec{a} \perp \vec{c}$$: $$\vec{a} \cdot \vec{c} = 0$$.

$$ |\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + 4 = \frac{4}{9\sin^2\alpha} + 4 $$ $$ 27|\vec{c} - \vec{a}|^2 = 27\left(\frac{4}{9\sin^2\alpha} + 4\right) = \frac{12}{\sin^2\alpha} + 108 $$

Minimize over $$\alpha \in [0, \pi/3]$$.

$$\frac{12}{\sin^2\alpha}$$ is minimized when $$\sin^2\alpha$$ is maximized. On $$[0, \pi/3]$$, $$\sin\alpha$$ is increasing, so maximum at $$\alpha = \pi/3$$: $$\sin(\pi/3) = \frac{\sqrt{3}}{2}$$, $$\sin^2(\pi/3) = \frac{3}{4}$$.

$$ 27|\vec{c} - \vec{a}|^2_{\min} = \frac{12}{3/4} + 108 = 16 + 108 = 124 $$

The correct answer is Option (2): 124.