Let a line L pass through the origin and be perpendicular to the...

Given,

A line $L$ pass through the origin and be perpendicular to the lines {L}_{1}: \overset{\to}{r} = \left(\overset{^}{i} - 11\overset{^}{j} - 7\overset{^}{k}\right) + \lambda\left(\overset{^}{i} + 2\overset{^}{j} + 3\overset{^}{k}\right), \lambda \in \mathbb{R} and
{L}_{2}: \overset{\to}{r} = \left(- \overset{^}{i} + \overset{^}{k}\right) + \mu\left(2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}\right), \mu \in \mathbb{R}

So, direction ratio of line will be,
\left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1\end{cases}\right|$</p><p>$= \overset{^}{i}\left(- 4\right) - \overset{^}{j}\left(- 5\right) + \overset{^}{k}\left(- 2\right)$</p><p>$= - 4\overset{^}{i} + 5\overset{^}{j} - 2\overset{^}{k}$</p><p>Hence, the equation of line$L$which passes through origin will be,</p><p>$L: \overset{\to}{r} = \sigma ( - 4\overset{^}{i} + 5\overset{^}{j} - 2\overset{^}{k} )$</p><p>Now finding the intersection point$P$we get,</p><p>$1 + \lambda = - 4\sigma . . . . \left(1\right)$</p><p>$- 11 + 2\lambda = 5\sigma . . . . \left(2\right)$</p><p>$- 7 + 3\lambda = - 2\sigma . . . . \left(3\right)$</p><p>So, from equation \left(1\right) \text{ and } \left(3\right)$ we get, $1 + \lambda = - 14 + 6\lambda ⇒ \lambda = 3, \sigma = - 1$

Hence, point $P ≡ \left(4, - 5, 2\right)$

Now finding the point $Q$ which is foot of perpendicular from point $P\left(4, - 5, 2\right)$ on line {L}_{2}: \left(- \overset{^}{i} + \overset{^}{k}\right) + \mu\left(2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}\right)$we get,</p><p>Now any point on line${L}_{2}$will be$Q\left(\left(2\mu - 1\right)\overset{^}{i} + 2\mu\overset{^}{j} + \left(\mu + 1\right)\overset{^}{k}\right)$</p><p>Now direction ratio of$PQwill be \left(- 5 + 2\mu\right)\overset{^}{i} + \left(2\mu + 5\right)\overset{^}{j} + \left(\mu - 1\right)\overset{^}{k} = 0

Now using the perpendicular condition we get,

$2 ( - 5 + 2\mu ) + 2 ( 2\mu + 5 ) + 1 ( \mu - 1 ) = 0$

$⇒ \mu = \dfrac{1}{9}$

Hence, $\alpha + \beta + \gamma = 2\mu - 1 + 2\mu + \mu + 1 = 5\mu = \dfrac{5}{9}$

$⇒ 9\left(\alpha + \beta + \gamma\right) = 5$