Let → a , → b and → c be three non-zero vectors such that → b and →...

Given: $\overset{\to}{a} + 5\overset{\to}{b}$ is collinear with $\overset{\to}{c}, \overset{\to}{b} + 6\overset{\to}{c}$ is collinear with $\overset{\to}{a} and \overset{\to}{a} + \alpha\overset{\to}{b} + \beta\overset{\to}{c} = \overset{\to}{0}$.

$⇒ \overset{\to}{a} + 5\overset{\to}{b} = \lambda\overset{\to}{c} . . . \left(i\right)</p><p>$⇒ \overset{\to}{b} + 6\overset{\to}{c} = \mu\overset{\to}{a} . . . \left(ii\right)$</p><p>$⇒ \overset{\to}{a} = \lambda\overset{\to}{c} - 5\overset{\to}{b} = \dfrac{\overset{\to}{b} + 6\overset{\to}{c}}{\mu}$</p><p>$⇒ \lambda\overset{\to}{c} - 5\overset{\to}{b} = \dfrac{6}{\mu}\overset{\to}{c} + \dfrac{1}{\mu}\overset{\to}{b}$</p><p>Comparing both the sides,</p><p>$⇒ \mu = \dfrac{- 1}{5}, \lambda = - 30$</p><p>Putting these values in equation \left(i\right)$

$⇒ \overset{\to}{a} + 5\overset{\to}{b} = - 30\overset{\to}{c}$

$⇒ \overset{\to}{a} + 5\overset{\to}{b} + 30\overset{\to}{c} = 0$

Comparing this equation with $\overset{\to}{a} + \alpha\overset{\to}{b} + \beta\overset{\to}{c} = \overset{\to}{0}$.

$⇒ \alpha = 5, \beta = 30$

$⇒ \alpha + \beta = 35$