If the points with position vectors $(\alpha \hat{i}+10 \hat{j}+13 \hat{k})$, $(6 \hat{i}+11 \hat{j}+11…
If the points with position vectors $(\alpha \hat{i}+10 \hat{j}+13 \hat{k})$, $(6 \hat{i}+11 \hat{j}+11 \hat{k}),\left(\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}\right)$ are collinear then $(19 \alpha-6 \beta)^2=$
$16$
$36$
$25$
$49$
Solution
Since, $(\alpha \hat{i}+10 \hat{j}+13 \hat{k}),(b \hat{i}+11 \hat{j}+11 \hat{k})$ and $\left(\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k}\right)$ are collinear.
So, $\Rightarrow \frac{\alpha-6}{\frac{3}{2}}=\frac{-1}{11-\beta}=\frac{2}{19} \frac{\alpha-6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta}=\frac{13-11}{11-(-8)}$
Now, $\alpha-6=\frac{3}{19}$ and $22-2 \beta=-19$ $\Rightarrow \alpha=6+\frac{9}{19}$ and $2 \beta=19+22$
$\begin{aligned} & \Rightarrow \alpha=6+\frac{9}{19} \text { and } 2 \beta=19+22 \\ & \Rightarrow \alpha=\frac{117}{19}, \text { and } 6 \beta=3 \times 41=123\end{aligned}$
$\begin{aligned} & \Rightarrow 19 \alpha=117 \text { and } 6 \beta=123 \\ & \text { So, }(19 \alpha-6 \beta)^2=(117-123)^2=(-6)^2=36\end{aligned}$