If the points with position vectors α ^ i + 10 ^ j + 13 ^ k , 6 ^ i...

The given position vectors can be written as

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

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,

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

$⇒ 11k + 11 = - 8k + 13$

$⇒ 19k = 2$

$⇒ k = \dfrac{2}{19}$

$∴$Ratio$= 2: 19$

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

$⇒ 19\alpha = 117$

$⇒ \alpha = \dfrac{117}{19}$

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

$⇒ \beta = \dfrac{41}{2}$

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

Hence this is the correct option.