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

We need to find $$(19\alpha - 6\beta)^2$$ given that three 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. Denote these points by $$A=(\alpha,10,13)$$, $$B=(6,11,11)$$ and $$C=\left(\frac{9}{2},\beta,-8\right)$$. Since collinear points imply parallel vectors, we compute $$\vec{AB}=(6-\alpha,1,-2)$$ and $$\vec{AC}=\left(\frac{9}{2}-\alpha,\beta-10,-21\right)$$. The proportionality of their components gives
$$\frac{6-\alpha}{\frac{9}{2}-\alpha}=\frac{1}{\beta-10}=\frac{-2}{-21}=\frac{2}{21}\,. $$ From $$\frac{1}{\beta-10}=\frac{2}{21}$$ we get $$\beta-10=\frac{21}{2}$$ and hence $$\beta=\frac{41}{2}$$. Using $$\frac{6-\alpha}{\frac{9}{2}-\alpha}=\frac{2}{21}$$ leads to
$$21(6-\alpha)=2\left(\frac{9}{2}-\alpha\right)\,,\quad 126-21\alpha=9-2\alpha\,,\quad 117=19\alpha\implies\alpha=\frac{117}{19}\,. $$

Finally, we compute $$19\alpha-6\beta=117-6\cdot\frac{41}{2}=117-123=-6$$ so that $$(19\alpha-6\beta)^2=(-6)^2=36\,. $$ The correct answer is Option A: 36.