up vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ as follows:

Let point $ A $ have the position vector $ \alpha \hat{\mathbf{i}} + 10 \hat{\mathbf{j}} + 13 \hat{\mathbf{k}} $.

Let point $ B $ have the position vector $ 6 \hat{\mathbf{i}} + 11 \hat{\mathbf{j}} + 11 \hat{\mathbf{k}} $.

Let point $ C $ have the position vector $ \frac{9}{2} \hat{\mathbf{i}} + \beta \hat{\mathbf{j}} - 8 \hat{\mathbf{k}} $.

The vector $ \overrightarrow{AB} $ is calculated as:

$ \overrightarrow{AB} = (6 - \alpha) \hat{\mathbf{i}} + (11 - 10) \hat{\mathbf{j}} + (11 - 13) \hat{\mathbf{k}} $

So:

$ \overrightarrow{AB} = (6 - \alpha) \hat{\mathbf{i}} + \hat{\mathbf{j}} - 2 \hat{\mathbf{k}} $

The vector $ \overrightarrow{AC} $ is calculated as:

$ \overrightarrow{AC} = \left(\frac{9}{2} - \alpha\right) \hat{\mathbf{i}} + (\beta - 10) \hat{\mathbf{j}} + (-8 - 13) \hat{\mathbf{k}} $

So:

$ \overrightarrow{AC} = \left(\frac{9}{2} - \alpha\right) \hat{\mathbf{i}} + (\beta - 10) \hat{\mathbf{j}} - 21 \hat{\mathbf{k}} $

Since the points are collinear, vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ are parallel, which implies:

$ \frac{6 - \alpha}{\frac{9}{2} - \alpha} = \frac{1}{\beta - 10} = \frac{-2}{-21} $

Solving these equations, we find:

$ \frac{1}{\beta - 10} = \frac{2}{21} \; \Rightarrow \; \beta - 10 = \frac{21}{2} $

$ \beta = 10 + \frac{21}{2} = \frac{41}{2} $

For the other ratio:

$ \frac{12 - 2\alpha}{9 - 2\alpha} = \frac{2}{21} $

$ 126 - 21\alpha = 9 - 2\alpha $

So:

$ 19\alpha = 117 $

$ \alpha = \frac{117}{19} = 6 $

Given:

$ (19\alpha - 6\beta)^2 = (117 - 123)^2 = (-6)^2 = 36 $

Thus, the final result is $(19\alpha - 6\beta)^2 = 36$.

Three Dimensional Geometry

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