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

$$\begin{gathered} \overline{\mathrm{OA}}=\mathrm{\alpha }\widehat{\mathrm{i}}+10\widehat{\mathrm{j}}+13\widehat{\mathrm{k}}, \overline{OB}=6\widehat{\mathrm{i}}+11\widehat{\mathrm{j}}+11\widehat{\mathrm{k}},\overline{\text{OC}}\text{\hspace{0.17em}=}\frac{9}{2}\widehat{\mathrm{i}}+\mathrm{\beta }\widehat{\mathrm{j}}-8\widehat{\mathrm{k}} \\ \Rightarrow \overline{AB}=\left(6-\alpha \right)\hat{i}+\hat{j}-2\hat{k} \\ \Rightarrow \overline{BC}=-\frac{3}{2}\hat{i}+\left(\beta -11\right)\hat{j}-19\hat{k} \end{gathered}$$
$\begin{array}{l}\frac{6-\alpha }{-\frac{3}{2}}=\frac{1}{\mathrm{\beta }-11}=\frac{-2}{-19}\Rightarrow 19\mathrm{\alpha }=19\left(6\right)+3=117\\ \Rightarrow 2\mathrm{\beta }=19+22=41\\ \therefore 19\mathrm{\alpha }-6\mathrm{\beta }=-6\end{array}$
$\therefore (19\alpha -6\beta {)}^2=(-6{)}^2=36$