^ If α=1+Σ_r=1^6(-3)^r-1 ^12 C_2 r-1, then the distance of the point (12, √3) from the line α x-√3 y+1=0 is .

$\displaystyle \begin{aligned} & \alpha=1+\sum_{\mathrm{r}=1}^6(-1)^{\mathrm{r}-1}{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} 3^{\mathrm{r}-1} \\ & \alpha=1+\sum_{\mathrm{r}=1}^6{ }^{12} \mathrm{C}_{2 \mathrm{r}-1} \frac{(\sqrt{3} \mathrm{i})^{2 \mathrm{t}-1}}{\sqrt{3} \mathrm{i}} \quad \mathrm{i}=\text { iota, let } \sqrt{3} \mathrm{i}=\mathrm{x}\end{aligned}$​α=1+r=1∑6​(−1)r−112C2r−1​3r−1α=1+r=1∑6​12C2r−1​3​i(3​i)2t−1​i= iota, let 3​i=x​ $\begin{aligned} \alpha & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left({ }^{12} \mathrm{C}_1 \mathrm{x}+{ }^{12} \mathrm{C}_3 \mathrm{x}^3+\ldots .{ }^{12} \mathrm{C}_{11} \mathrm{x}^{11}\right) \\ & =1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{(1+\sqrt{3} \mathrm{i})^{12}-(1-\sqrt{3} \mathrm{i})^{12}}{2}\right)\end{aligned}$α​=1+3​i1​(12C1​x+12C3​x3+….12C11​x11)=1+3​i1​(2(1+3​i)12−(1−3​i)12​)​ $=1+\frac{1}{\sqrt{3} \mathrm{i}}\left(\frac{\left(-2 \mathrm{w}^2\right)^{12}-(2 \mathrm{w})^{12}}{2}\right)=1$=1+3​i1​(2(−2w2)12−(2w)12​)=1 so distance of $(12, \sqrt{3})$(12,3​) from $x-\sqrt{3} y+1=0$x−3​y+1=0 is $\frac{12-3+1}{2}=5$212−3+1​=5