Let the line of the shortest distance between the lines L _ 1 : → r...

Let point on ${L}_{1}$ be $P$\left(\left(1 + \lambda\right)\overset{^}{i} + \left(2 - \lambda\right)\overset{^}{j} + \left(3 + \lambda\right)\overset{^}{k}\right) and point on ${L}_{2}$ be Q\left(\left(4 + \mu\right)\overset{^}{i} + \left(5 + \mu\right)\overset{^}{j} + \left(6 - \mu\right)\overset{^}{k}\right).

⇒ \overset{\to}{PQ} = \left(\lambda - \mu - 3\right)\overset{^}{i} + \left(- \lambda - \mu - 3\right)\overset{^}{j} + \left(\lambda + \mu - 3\right)\overset{^}{k}

Now, $\overset{\to}{PQ} ⊥ {L}_{1}, {L}_{2}$

$⇒ \left(\lambda - \mu - 3\right) + \left(\lambda + \mu + 3\right) + \left(\lambda + \mu - 3\right) = 0$

$⇒ 3\lambda + \mu = 3 . . . \left(i\right)$

Also, \left(\lambda - \mu - 3\right) + \left(- \lambda - \mu - 3\right) + \left(- \lambda - \mu + 3\right) = 0$</p><p>$⇒ - \lambda - 3\mu = 3 . . . \left(ii\right)$</p><p>Now, solving above equations we get,</p><p>$⇒ 3\lambda + \mu - 3\lambda - 9\mu = 3 + 9$</p><p>$⇒ - 8\mu = 12$</p><p>$⇒ \mu = \dfrac{- 3}{2}$</p><p>$⇒ \lambda = \dfrac{3}{2}$</p><p>$⇒ P ≡ \left(\dfrac{5}{2}, \dfrac{1}{2}, \dfrac{9}{2}\right), Q ≡ \left(\dfrac{5}{2}, \dfrac{7}{2}, \dfrac{15}{2}\right)$</p><p>So, mid-point of$PQ$is \left(\dfrac{5}{2}, 2, 6\right)$

$⇒ 2\left(\alpha + \beta + \gamma\right) = 2\left(\dfrac{5}{2} + 2 + 6\right)$

$⇒ 2\left(\alpha + \beta + \gamma\right) = 21$