Let $S$ be the set of all values of $\lambda$, for which the shortest distance between the lines $\frac{x-\lambda }{0}=\frac{y-3}{4}=\frac{z+6}{1}$and $\frac{x+\lambda }{3}=\frac{y}{-4}=\frac{z-6}{0}$ is $13$. Then $8\left|\sum _{\lambda \in S}\lambda \right|$ is equal to 

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

We know that the shortest distance between two lines is $d=\left|\frac{\left({\overrightarrow{a}}_2-{\overrightarrow{a}}_1\right)·\left({\overrightarrow{n}}_1\times {\overrightarrow{n}}_2\right)}{\left|{\overrightarrow{n}}_1\times {\overrightarrow{n}}_2\right|}\right| \& d=13 \left(\text{given}\right)$

$\Rightarrow {\overrightarrow{n}}_1\times {\overrightarrow{n}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 4& 1\\ 3& -4& 0\end{array}\right|$$=4\hat{i}+3\hat{j}-12\hat{k}$

Now using the above formula of distance we get,

$\Rightarrow \left|\frac{(2\lambda \hat{i}+3\hat{j}-12\hat{k})·(4\hat{i}+3\hat{j}-12\hat{k})}{\sqrt{16+9+144}}\right|=13$

$\Rightarrow \left|\frac{(2\lambda \hat{i}+3\hat{j}-12\hat{k})·(4\hat{i}+3\hat{j}-12\hat{k})}{13}\right|=13$

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

$\Rightarrow \left|8\lambda +153\right|=169$

$\Rightarrow 8\lambda =\pm 169-153$

$\Rightarrow \lambda =\frac{16}{8}, \frac{-322}{8}$

$\Rightarrow 8\left|\sum _{\lambda \in s}\lambda \right|=8\left|\frac{16}{8}-\frac{322}{8}\right|=306$

Hence this is the correct option.