If $$\alpha =1+\sum_{r=1}^{6}(-3)^{r-1}$$ $$^{12}C_{2r-1}$$, then the distance of the point $$(12,\sqrt{3})$$ from the line $$\alpha x-\sqrt{3}y+1$$ is________

Find the value of $$\alpha$$

We use the binomial expansion of $$(1 + i\sqrt{3})^{12}$$:

$$(1 + i\sqrt{3})^{12} = \binom{12}{0} + \binom{12}{1}(i\sqrt{3}) + \binom{12}{2}(i\sqrt{3})^2 + \binom{12}{3}(i\sqrt{3})^3 + \dots$$

The imaginary part is:

$$\text{Im}[(1 + i\sqrt{3})^{12}] = \binom{12}{1}(\sqrt{3}) - \binom{12}{3}(\sqrt{3})^3 + \binom{12}{5}(\sqrt{3})^5 - \dots$$

Divide by $$\sqrt{3}$$:

$$\frac{\text{Im}[(1 + i\sqrt{3})^{12}]}{\sqrt{3}} = \binom{12}{1} - 3\binom{12}{3} + 3^2\binom{12}{5} - \dots = \sum_{r=1}^{6}(-3)^{r-1} \binom{12}{2r-1}$$

Using De Moivre's Theorem:

$$1 + i\sqrt{3} = 2(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3})$$

$$(1 + i\sqrt{3})^{12} = 2^{12}(\cos 4\pi + i\sin 4\pi) = 4096(1 + 0i)$$

Since the imaginary part is $$0$$, the summation equals $$0$$.

Therefore, $$\alpha = 1 + 0 = \mathbf{1}$$.

Calculate the distance

The line is $$x - \sqrt{3}y + 1 = 0$$. Distance from $$(12, \sqrt{3})$$:

$$d = \frac{|12 - \sqrt{3}(\sqrt{3}) + 1|}{\sqrt{1^2 + (-\sqrt{3})^2}} = \frac{|12 - 3 + 1|}{\sqrt{4}} = \frac{10}{2} = \mathbf{5}$$