QJEEM25Ma6f6b: JEE Main 2025 Mathematics - Class 11 JEEM25Ma6f6b Solution: If , then the distance of the point from the line is ________.

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