Let S be the set of all values of λ , for which the shortest...

The given lines are$\dfrac{x - \lambda}{0} = \dfrac{y - 3}{4} = \dfrac{z + 6}{1}$and $\dfrac{x + \lambda}{3} = \dfrac{y}{- 4} = \dfrac{z - 6}{0}$

We know that the shortest distance between two lines is $d = \left|\dfrac{\left(\overset{\to}{a}2 - \overset{\to}{a}1\right) \cdot \left(\overset{\to}{n}1 \times \overset{\to}{n}2\right)}{\left|\overset{\to}{n}1 \times \overset{\to}{n}2\right|}\right| \text{ and } d = 13\left(\text{given}\right)$

⇒ \overset{\to}{n}1 \times \overset{\to}{n}2 = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 0 & 4 & 1 \\ 3 & - 4 & 0\end{cases}\right|= 4\overset{^}{i} + 3\overset{^}{j} - 12\overset{^}{k}

Now using the above formula of distance we get,

⇒ \left|\dfrac{( 2\lambda\overset{^}{i} + 3\overset{^}{j} - 12\overset{^}{k} ) \cdot ( 4\overset{^}{i} + 3\overset{^}{j} - 12\overset{^}{k} )}{\sqrt{16 + 9 + 144}}\right| = 13

⇒ \left|\dfrac{( 2\lambda\overset{^}{i} + 3\overset{^}{j} - 12\overset{^}{k} ) \cdot ( 4\overset{^}{i} + 3\overset{^}{j} - 12\overset{^}{k} )}{13}\right| = 13

$⇒ \left|8\lambda + 9 + 144\right| = 169$

$⇒ \left|8\lambda + 153\right| = 169$

$⇒ 8\lambda = \pm 169 - 153$

$⇒ \lambda = \dfrac{16}{8}, \dfrac{- 322}{8}$

$⇒ 8\left|\mathop{∑}_{\lambda \in s}\lambda\right| = 8\left|\dfrac{16}{8} - \dfrac{322}{8}\right| = 306$

Hence this is the correct option.