Let $$\hat{a}, \hat{b}$$ be unit vectors. If $$\vec{c}$$ be a vector such that the angle between $$\hat{a}$$ and $$\vec{c}$$ is $$\frac{\pi}{12}$$, and $$\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})$$, then $$|6\vec{c}|^2$$ is equal to:

Let $$\hat{a}$$ and $$\hat{b}$$ be unit vectors, and $$\vec{c}$$ be a vector such that the angle between $$\hat{a}$$ and $$\vec{c}$$ is $$\frac{\pi}{12}$$, and $$\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})$$.

Take the magnitude squared of both sides.

$$|\hat{b}|^2 = |\vec{c} + 2(\vec{c} \times \hat{a})|^2$$

$$1 = |\vec{c}|^2 + 4|\vec{c} \times \hat{a}|^2 + 4\vec{c} \cdot (\vec{c} \times \hat{a})$$

Simplify the cross terms.

The scalar triple product $$\vec{c} \cdot (\vec{c} \times \hat{a}) = 0$$ (since it involves two identical vectors in the triple product).

$$|\vec{c} \times \hat{a}|^2 = |\vec{c}|^2 \sin^2\left(\frac{\pi}{12}\right)$$

Substitute and simplify.

$$1 = |\vec{c}|^2 + 4|\vec{c}|^2 \sin^2\left(\frac{\pi}{12}\right) = |\vec{c}|^2\left(1 + 4\sin^2\frac{\pi}{12}\right)$$

Compute $$\sin^2\left(\frac{\pi}{12}\right)$$.

Using the identity $$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$$:

$$\sin^2\frac{\pi}{12} = \frac{1 - \cos\frac{\pi}{6}}{2} = \frac{1 - \frac{\sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}$$

Find the coefficient.

$$1 + 4 \cdot \frac{2 - \sqrt{3}}{4} = 1 + 2 - \sqrt{3} = 3 - \sqrt{3}$$

$$|\vec{c}|^2 = \frac{1}{3 - \sqrt{3}}$$

Compute $$|6\vec{c}|^2$$.

$$|6\vec{c}|^2 = 36|\vec{c}|^2 = \frac{36}{3 - \sqrt{3}}$$

Rationalizing: $$= \frac{36(3 + \sqrt{3})}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{36(3 + \sqrt{3})}{9 - 3} = \frac{36(3 + \sqrt{3})}{6} = 6(3 + \sqrt{3})$$

Answer: Option B