Key Concepts and Formulas

• Collinearity of Three Points in 3D: Three points $A, B, C$ with position vectors $\vec{a}, \vec{b}, \vec{c}$ are collinear if the vector $\vec{AB}$ is parallel to $\vec{BC}$, which means $\vec{AB} = k \vec{BC}$ for some scalar $k$. This implies that the ratios of the differences in their corresponding coordinates are equal: $\frac{x_A-x_B}{x_B-x_C} = \frac{y_A-y_B}{y_B-y_C} = \frac{z_A-z_B}{z_B-z_C}$
• Algebraic Manipulation: Solving linear equations and simplifying expressions involving fractions and variables.

Step-by-Step Solution

Step 1: Identify the position vectors and coordinates of the given points. Let the three given points be $P_1$, $P_2$, and $P_3$ with position vectors: $\vec{p_1} = \alpha \hat{i}+10 \hat{j}+13 \hat{k} \implies P_1(\alpha, 10, 13)$ $\vec{p_2} = 6 \hat{i}+11 \hat{j}+11 \hat{k} \implies P_2(6, 11, 11)$ $\vec{p_3} = \frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k} \implies P_3\left(\frac{9}{2}, \beta, -8\right)$

Step 2: Apply the condition for collinearity using the ratios of coordinate differences. Since the points are collinear, the ratios of the differences in their corresponding coordinates must be equal. We will use the form $\frac{x_1-x_2}{x_2-x_3} = \frac{y_1-y_2}{y_2-y_3} = \frac{z_1-z_2}{z_2-z_3}$.

• x-coordinates: $x_1 - x_2 = \alpha - 6$ $x_2 - x_3 = 6 - \frac{9}{2} = \frac{12 - 9}{2} = \frac{3}{2}$ Ratio for x: $\frac{\alpha - 6}{3/2} = \frac{2(\alpha - 6)}{3}$

• y-coordinates: $y_1 - y_2 = 10 - 11 = -1$ $y_2 - y_3 = 11 - \beta$ Ratio for y: $\frac{-1}{11 - \beta}$

• z-coordinates: $z_1 - z_2 = 13 - 11 = 2$ $z_2 - z_3 = 11 - (-8) = 11 + 8 = 19$ Ratio for z: $\frac{2}{19}$

Equating these ratios, we get: $\frac{2(\alpha - 6)}{3} = \frac{-1}{11 - \beta} = \frac{2}{19}$

Step 3: Solve for $\alpha$ by equating the x-ratio and the z-ratio. We use the known numerical ratio $\frac{2}{19}$ to find the values of $\alpha$ and $\beta$. $\frac{2(\alpha - 6)}{3} = \frac{2}{19}$ Multiply both sides by 3 and divide by 2: $\alpha - 6 = \frac{2}{19} \times \frac{3}{2}$ $\alpha - 6 = \frac{3}{19}$ Add 6 to both sides: $\alpha = 6 + \frac{3}{19} = \frac{6 \times 19 + 3}{19} = \frac{114 + 3}{19} = \frac{117}{19}$

Step 4: Solve for $\beta$ by equating the y-ratio and the z-ratio. $\frac{-1}{11 - \beta} = \frac{2}{19}$ Cross-multiply: $-1 \times 19 = 2 \times (11 - \beta)$ $-19 = 22 - 2\beta$ Rearrange the terms to solve for $\beta$: $2\beta = 22 + 19$ $2\beta = 41$ $\beta = \frac{41}{2}$

Step 5: Calculate the value of the expression $(19 \alpha - 6 \beta)^2$. Substitute the found values of $\alpha$ and $\beta$ into the expression: $(19 \alpha - 6 \beta)^2 = \left( 19 \times \frac{117}{19} - 6 \times \frac{41}{2} \right)^2$ Simplify the terms inside the parentheses: $19 \times \frac{117}{19} = 117$ $6 \times \frac{41}{2} = 3 \times 41 = 123$ Substitute these values back into the expression: $(117 - 123)^2 = (-6)^2 = 36$

Common Mistakes & Tips

• Order of Subtraction: Be consistent with the order of subtraction when forming the coordinate differences (e.g., always $P_1 - P_2$ and $P_2 - P_3$). Mixing the order will lead to incorrect signs and ratios.
• Fraction Arithmetic: Double-check all calculations involving fractions, especially when adding, subtracting, or cross-multiplying.
• Sign Errors: Carefully handle negative signs, particularly when dealing with subtraction of coordinates or terms in equations.

Summary

The problem requires us to use the condition of collinearity for three points in 3D space, which states that the ratios of the differences in their corresponding coordinates are equal. By setting up these ratios using the given position vectors, we derived two equations involving the unknowns $\alpha$ and $\beta$. Solving these equations allowed us to find the specific values of $\alpha$ and $\beta$. Finally, we substituted these values into the expression $(19 \alpha - 6 \beta)^2$ to compute the required result.

The final answer is $\boxed{\text{36}}$.