If alpha = 1 + sum_r=1^6 (-3)^r-1 ^12 C_2r-1$\alpha = 1 + \sum_{r=1}^{6} (-3)^{r-1} {}^{12} C_{2r-1}$, then the distance of the point (12, sqrt3)$(12, \sqrt{3})$ from the line alpha x - sqrt3 y + 1 = 0$\alpha x - \sqrt{3} y + 1 = 0$ is

### Related Formula Perpendicular distance from a point (x_1, y_1)$(x_1, y_1)$ to a line Ax + By + C = 0$Ax + By + C = 0$: d = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$ ### Core Logic Rewrite the summation for alpha$\alpha$ by embedding complex roots to isolate alternating terms: alpha = 1 + frac1sqrt3i left[ frac(1 + sqrt3i)^12 - (1 - sqrt3i)^122 right]$$\alpha = 1 + \frac{1}{\sqrt{3}i} \left[ \frac{(1 + \sqrt{3}i)^{12} - (1 - \sqrt{3}i)^{12}}{2} \right]$$ Using complex cube roots of unity conversions (1 + sqrt3i = -2omega^2$1 + \sqrt{3}i = -2\omega^2$), this simplifies directly to alpha = 1$\alpha = 1$. ### Step 1: Calculating Perpendicular Distance Substitute alpha = 1$\alpha = 1$ into the line equation, yielding x - sqrt3y + 1 = 0$x - \sqrt{3}y + 1 = 0$. Evaluating the distance from point (12, sqrt3)$(12, \sqrt{3})$: d = frac|12 - sqrt3(sqrt3) + 1|sqrt1^2 + (-sqrt3)^2 = frac|12 - 3 + 1|2 = frac102 = 5$$d = \frac{|12 - \sqrt{3}(\sqrt{3}) + 1|}{\sqrt{1^2 + (-\sqrt{3})^2}} = \frac{|12 - 3 + 1|}{2} = \frac{10}{2} = 5$$ ### Pattern Recognition Binomial series containing terms like sqrt3^2r-1$\sqrt{3}^{2r-1}$ collapse neatly into simple geometric forms using standard complex rotations (e^ipi/3$e^{i\pi/3}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Binomial Theorem Class 11 Maths: Straight Lines