For $$p > 0$$, a vector $$\vec{v_2} = 2\hat{i} + (p+1)\hat{j}$$ is obtained by rotating the vector $$\vec{v_1} = \sqrt{3}p\hat{i} + \hat{j}$$ by an angle $$\theta$$ about origin in counter clockwise direction. If $$\tan\theta = \frac{(\alpha\sqrt{3}-2)}{(4\sqrt{3}+3)}$$, then the value of $$\alpha$$ is equal to ___.

We have $$\vec{v_1} = \sqrt{3}p\hat{i} + \hat{j}$$ and $$\vec{v_2} = 2\hat{i} + (p+1)\hat{j}$$. Since $$\vec{v_2}$$ is obtained by rotating $$\vec{v_1}$$, their magnitudes must be equal: $$|\vec{v_1}|^2 = 3p^2 + 1 = |\vec{v_2}|^2 = 4 + (p+1)^2 = 4 + p^2 + 2p + 1 = p^2 + 2p + 5$$

$$3p^2 + 1 = p^2 + 2p + 5 \implies 2p^2 - 2p - 4 = 0 \implies p^2 - p - 2 = 0 \implies (p-2)(p+1) = 0$$

Since $$p > 0$$, we have $$p = 2$$. So $$\vec{v_1} = 2\sqrt{3}\hat{i} + \hat{j}$$ and $$\vec{v_2} = 2\hat{i} + 3\hat{j}$$.

The angle $$\theta$$ from $$\vec{v_1}$$ to $$\vec{v_2}$$ (counterclockwise) satisfies: $$\tan\theta = \tan(\phi_2 - \phi_1)$$ where $$\phi_1 = \arg(\vec{v_1})$$ and $$\phi_2 = \arg(\vec{v_2})$$.

Using the cross product and dot product: $$\sin\theta = \frac{v_{1x} v_{2y} - v_{1y} v_{2x}}{|\vec{v_1}||\vec{v_2}|} = \frac{2\sqrt{3} \cdot 3 - 1 \cdot 2}{|\vec{v_1}||\vec{v_2}|} = \frac{6\sqrt{3} - 2}{|\vec{v_1}||\vec{v_2}|}$$

$$\cos\theta = \frac{\vec{v_1}\cdot\vec{v_2}}{|\vec{v_1}||\vec{v_2}|} = \frac{2\sqrt{3}\cdot 2 + 1\cdot 3}{|\vec{v_1}||\vec{v_2}|} = \frac{4\sqrt{3}+3}{|\vec{v_1}||\vec{v_2}|}$$

Therefore: $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{6\sqrt{3}-2}{4\sqrt{3}+3}$$

Comparing with $$\tan\theta = \frac{\alpha\sqrt{3}-2}{4\sqrt{3}+3}$$, we get $$\alpha\sqrt{3} = 6\sqrt{3}$$, so $$\alpha = 6$$.