If $\alpha=1+\sum\limits_{r=1}^6(-3)^{r-1} \quad{ }^{12} \mathrm{C}_{2 r-1}$, then the distance of the point $(12, \sqrt{3})$ from the line $\alpha x-\sqrt{3} y+1=0$ is ________.

<p>$$\begin{aligned} &\begin{aligned} \alpha & =1+\sum_{\mathrm{r}=1}^6(-1)^{\mathrm{r}-1}{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} 3^{\mathrm{r}-1} \\ \alpha & =1+\sum_{\mathrm{r}=1}^6{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} \frac{(\sqrt{3} \mathrm{i})^{2 \mathrm{t}-1}}{\sqrt{3} \mathrm{i}} \quad \mathrm{i}=\text { iota, let } \sqrt{3} \mathrm{i}=\mathrm{x} \\ \alpha & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left({ }^{12} \mathrm{C}_1 \mathrm{x}+{ }^{12} \mathrm{C}_3 \mathrm{x}^3+\ldots .{ }^{12} \mathrm{C}_{11} \mathrm{x}^{11}\right) \\ & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{(1+\sqrt{3} \mathrm{i})^{12}-(1-\sqrt{3} \mathrm{i})^{12}}{2}\right) \\ & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{\left(-2 \omega^2\right)^{12}-(2 \omega)^{12}}{2}\right)=1 \end{aligned}\\ &\text { so distance of }(12, \sqrt{3}) \text { from } x-\sqrt{3} y+1=0 \text { is }\\ &\frac{12-3+1}{2}=5 \end{aligned}$$</p>