Given : Points with position vectors

$$ \alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k} $$

and $\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$ are collinear.

So, $\frac{\alpha-6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta}=\frac{13-11}{11+8}$

$$ \begin{aligned} & \Rightarrow \frac{2(\alpha-6)}{3}=\frac{-1}{11-\beta}=\frac{2}{19} \\\\ & \Rightarrow \frac{2}{3}(\alpha-6)=\frac{2}{19} \end{aligned} $$

$$ \begin{aligned} & \Rightarrow 19 \alpha-114=3 \Rightarrow 19 \alpha=117 \\\\ & \Rightarrow \alpha=\frac{117}{19} \end{aligned} $$

And, $\frac{-1}{11-\beta}=\frac{2}{19}$

$$ \begin{aligned} & \Rightarrow-19=22-2 \beta \\\\ & \Rightarrow 2 \beta=41 \\\\ & \Rightarrow \beta=\frac{41}{2} \end{aligned} $$

$$ \begin{aligned} & \therefore(19 \alpha-6 \beta)^2=\left(19 \times \frac{117}{19}-\frac{6 \times 41}{2}\right)^2 \\\\ & =(117-123)^2=36 \end{aligned} $$