If the points with position vectors $$\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}, \frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$$ are collinear, then $(19 \alpha-6 \beta)^{2}$ is equal to :

Given : Points with position vectors <br/><br/>$\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}$ <br/><br/>and $\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}$ are collinear. <br/><br/>So, $\frac{\alpha-6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta}=\frac{13-11}{11+8}$ <br/><br/>$$ \begin{aligned} & \Rightarrow \frac{2(\alpha-6)}{3}=\frac{-1}{11-\beta}=\frac{2}{19} \\\\ & \Rightarrow \frac{2}{3}(\alpha-6)=\frac{2}{19} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow 19 \alpha-114=3 \Rightarrow 19 \alpha=117 \\\\ & \Rightarrow \alpha=\frac{117}{19} \end{aligned} $$ <br/><br/>And, $\frac{-1}{11-\beta}=\frac{2}{19}$ <br/><br/>$$ \begin{aligned} & \Rightarrow-19=22-2 \beta \\\\ & \Rightarrow 2 \beta=41 \\\\ & \Rightarrow \beta=\frac{41}{2} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \therefore(19 \alpha-6 \beta)^2=\left(19 \times \frac{117}{19}-\frac{6 \times 41}{2}\right)^2 \\\\ & =(117-123)^2=36 \end{aligned} $$