If the points with position vectors α i ^ + 10 j ^ + 13 k ^ ,   6 i ^ + 11 j ^ + 11 k ^ , 9 2 i ^…

The given position vectors can be written as

• $A(\alpha , 10, 13)$
• $B(6, 11, 11)$
• $C\left(\frac{9}{2}, \beta , -8\right)$

Also given that these three points are collinear.

Let us assume that the point $B$ divides $AB, BC$ in the ratio $k:1$.

Since, $A, B, C$ are collinear 

On applying section formula for $z$ co-ordinate we get,

$11=\frac{-8k+13}{k+1}$

$\Rightarrow 11k+11=-8k+13$

$\Rightarrow 19k= 2$

$\Rightarrow k=\frac{2}{19}$

$\therefore$Ratio$=2 : 19$

Now, $\frac{\alpha \times 19+\left({\displaystyle \frac{9}{2}}\right)\times 2}{2+19}=6$

$\Rightarrow 19\alpha =117$

$\Rightarrow \alpha =\frac{117}{19}$

Now similarly,  $\frac{2\beta +190}{21}=11$

$\Rightarrow \beta =\frac{41}{2}$

$\therefore {\left(19\alpha -6\beta \right)}^2={\left(117-123\right)}^2=36$

Hence this is the correct option.