Key Concepts Utilized

This problem requires the application of several key mathematical concepts:

1. Binomial Theorem for Sums of Odd/Even Terms: Specifically, the property that relates sums of binomial coefficients with odd or even lower indices to the expansion of $(1+x)^n$ and $(1-x)^n$. The general identity used is: $\sum_{k \text{ odd}} \binom{n}{k} x^k = \frac{(1+x)^n - (1-x)^n}{2}$ and if we need terms like $\binom{n}{1} + \binom{n}{3}y + \binom{n}{5}y^2 + \dots$, we can substitute $x^2 = y$ (or $x = \sqrt{y}$ or $x = i\sqrt{-y}$).
2. Complex Numbers and De Moivre's Theorem: When dealing with terms like $\sqrt{-3}$ or evaluating powers of complex numbers, converting them to polar form and applying De Moivre's Theorem, $(r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))$, greatly simplifies calculations.
3. Properties of Cube Roots of Unity ($\omega$): Recognizing expressions like $1 \pm i\sqrt{3}$ as related to $2\omega$ or $2\omega^2$ can provide an alternative path for simplification, leveraging properties like $\omega^3=1$.
4. Distance of a Point from a Line: The formula for the perpendicular distance of a point $(x_0, y_0)$ from a line $Ax + By + C = 0$ is given by: $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

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Part 1: Evaluating the value of $\alpha$

The first step is to evaluate the given expression for $\alpha$: $\alpha = 1+\sum_{r=1}^6(-3)^{r-1} {}^{12} \mathrm{C}_{2 r-1}$

Let's expand the sum for better clarity: $\sum_{r=1}^6(-3)^{r-1} {}^{12} \mathrm{C}_{2 r-1} = {}^{12} \mathrm{C}_1 (-3)^0 + {}^{12} \mathrm{C}_3 (-3)^1 + {}^{12} \mathrm{C}_5 (-3)^2 + \ldots + {}^{12} \mathrm{C}_{11} (-3)^5$ This sum involves binomial coefficients with odd lower indices (1, 3, 5, ..., 11) and powers of $(-3)$. This structure suggests using the binomial expansion of $(1+x)^n - (1-x)^n$.

Step 1: Relating the sum to binomial expansion We know that for an integer $n$: $(1+x)^n = {}^{n} \mathrm{C}_0 + {}^{n} \mathrm{C}_1 x + {}^{n} \mathrm{C}_2 x^2 + {}^{n} \mathrm{C}_3 x^3 + \ldots + {}^{n} \mathrm{C}_n x^n$ $(1-x)^n = {}^{n} \mathrm{C}_0 - {}^{n} \mathrm{C}_1 x + {}^{n} \mathrm{C}_2 x^2 - {}^{n} \mathrm{C}_3 x^3 + \ldots + (-1)^n {}^{n} \mathrm{C}_n x^n$ Subtracting these two expansions: $(1+x)^n - (1-x)^n = 2 \left( {}^{n} \mathrm{C}_1 x + {}^{n} \mathrm{C}_3 x^3 + {}^{n} \mathrm{C}_5 x^5 + \ldots \right)$ In our problem, $n=12$. The sum is of the form ${}^{12} \mathrm{C}_1 + {}^{12} \mathrm{C}_3 (-3) + {}^{12} \mathrm{C}_5 (-3)^2 + \ldots$. To match the term $(-3)^{r-1}$ with $x^{2r-1}$ (or $x^1, x^3, x^5, \dots$), we need to find an $x$ such that $x^2 = -3$. Let $x = i\sqrt{3}$. Then $x^2 = (i\sqrt{3})^2 = i^2 (\sqrt{3})^2 = -1 \times 3 = -3$. So, we can rewrite the sum as: $\sum_{r=1}^6 (-3)^{r-1} {}^{12} \mathrm{C}_{2r-1} = \frac{1}{i\sqrt{3}} \sum_{r=1}^6 {}^{12} \mathrm{C}_{2r-1} (i\sqrt{3})^{2r-1}$ Comparing this to the identity for odd terms, with $n=12$ and $x=i\sqrt{3}$, we get: $\sum_{r=1}^6 {}^{12} \mathrm{C}_{2r-1} (i\sqrt{3})^{2r-1} = \frac{(1+i\sqrt{3})^{12} - (1-i\sqrt{3})^{12}}{2}$ Therefore, the sum simplifies to: $\sum_{r=1}^6 (-3)^{r-1} {}^{12} \mathrm{C}_{2r-1} = \frac{1}{i\sqrt{3}} \left( \frac{(1+i\sqrt{3})^{12} - (1-i\sqrt{3})^{12}}{2} \right)$

Step 2: Evaluating the complex number powers using cube roots of unity To evaluate $(1+i\sqrt{3})^{12}$ and $(1-i\sqrt{3})^{12}$, we can use properties of cube roots of unity. Recall that the complex cube roots of unity are $1, \omega, \omega^2$, where $\omega = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\omega^2 = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. Let's express $1+i\sqrt{3}$ and $1-i\sqrt{3}$ in terms of $\omega$ or $\omega^2$: $1+i\sqrt{3} = -2\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -2\omega^2$ $1-i\sqrt{3} = -2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -2\omega$ Now, substitute these into the expression: $\frac{1}{i\sqrt{3}} \left( \frac{(-2\omega^2)^{12} - (-2\omega)^{12}}{2} \right)$ Since $12$ is an even power, $(-2)^{12} = 2^{12}$. $= \frac{1}{i\sqrt{3}} \left( \frac{2^{12}(\omega^2)^{12} - 2^{12}\omega^{12}}{2} \right)$ $= \frac{1}{i\sqrt{3}} \left( \frac{2^{12}\omega^{24} - 2^{12}\omega^{12}}{2} \right)$ Using the property $\omega^3 = 1$: $\omega^{24} = (\omega^3)^8 = 1^8 = 1$ $\omega^{12} = (\omega^3)^4 = 1^4 = 1$ Substitute these values back: $= \frac{1}{i\sqrt{3}} \left( \frac{2^{12}(1) - 2^{12}(1)}{2} \right) = \frac{1}{i\sqrt{3}} \left( \frac{0}{2} \right) = 0$ So, the sum part evaluates to $0$. Therefore, $\alpha = 1 + 0 = 1$

Common Mistake to Avoid: A common pitfall here is incorrectly handling the powers of $i$ or the signs when substituting complex numbers. Always ensure careful expansion and application of De Moivre's Theorem or properties of roots of unity.

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Part 2: Calculating the distance of the point from the line

We have found $\alpha = 1$. The line equation is given as $\alpha x - \sqrt{3} y + 1 = 0$. Substituting $\alpha = 1$, the equation of the line becomes: $x - \sqrt{3} y + 1 = 0$ We need to find the distance of the point $(12, \sqrt{3})$ from this line.

Step 1: Identify coefficients and point coordinates For the line $Ax + By + C = 0$: $A = 1$ $B = -\sqrt{3}$ $C = 1$ For the point $(x_0, y_0)$: $x_0 = 12$ $y_0 = \sqrt{3}$

Step 2: Apply the distance formula The distance $D$ is given by: $D = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ Substitute the values: $D = \frac{|(1)(12) + (-\sqrt{3})(\sqrt{3}) + 1|}{\sqrt{(1)^2 + (-\sqrt{3})^2}}$ $D = \frac{|12 - 3 + 1|}{\sqrt{1 + 3}}$ $D = \frac{|10|}{\sqrt{4}}$ $D = \frac{10}{2}$ $D = 5$

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Tips for Success

• Always simplify expressions for binomial sums carefully, paying attention to the powers of $x$ and the coefficients.
• When dealing with complex numbers raised to powers, polar form and De Moivre's theorem are usually the most efficient methods. Recognizing relationships with roots of unity can also be a powerful shortcut.
• Remember the absolute value in the distance formula to ensure the distance is always non-negative.
• Double-check arithmetic, especially with signs and square roots.

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Summary

The problem required two main steps: first, to calculate the value of $\alpha$ using properties of binomial expansions and complex numbers, and then to find the distance of a given point from a line whose equation involved $\alpha$.

By recognizing the sum in the definition of $\alpha$ as related to the odd terms of a binomial expansion, we introduced $x = i\sqrt{3}$. Subsequently, we evaluated the powers of the complex numbers $(1+i\sqrt{3})^{12}$ and $(1-i\sqrt{3})^{12}$ by expressing them in terms of cube roots of unity ($\omega$ and $\omega^2$) and applying properties of $\omega^3=1$. This simplification revealed that the sum part was $0$, leading to $\alpha = 1$.

Finally, using the calculated value of $\alpha=1$, the line equation became $x - \sqrt{3} y + 1 = 0$. Applying the standard formula for the distance of a point $(12, \sqrt{3})$ from this line yielded a distance of $5$.