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 $$8|\sum_{\lambda \in S} \lambda|$$ is equal to

Line 1: $$\frac{x-\lambda}{0} = \frac{y-3}{-4} = \frac{z+6}{1} \implies \vec{a}_1 = (\lambda, 3, -6)$$, and direction vector $$\vec{b} = (0, -4, 1)$$

Line 2: $$\frac{x+\lambda}{3} = \frac{y}{-4} = \frac{z-6}{0} \implies \vec{a}_2 = (-\lambda, 0, 6)$$, and direction vector $$\vec{d} = (3, -4, 0)$$

$$\vec{a}_1 - \vec{a}_2 = (\lambda - (-\lambda), 3 - 0, -6 - 6) = (2\lambda, 3, -12)$$

$$\text{Shortest Distance} = \frac{\begin{vmatrix} 2\lambda & 3 & -12 \\ 0 & 4 & 1 \\ 3 & -4 & 0 \end{vmatrix}}{\sqrt{144 + 9 + 16}}$$

$$\begin{vmatrix} 2\lambda & 3 & -12 \\ 0 & 4 & 1 \\ 3 & -4 & 0 \end{vmatrix} = 2\lambda(0 - (-4)) - 3(0 - 3) - 12(0 - 12)$$

$$= 2\lambda(4) - 3(-3) - 12(-12) = 8\lambda + 9 + 144 = 8\lambda + 153$$

$$\frac{|8\lambda + 153|}{13} = 13 \implies |8\lambda + 153| = 169$$

$$8\lambda + 153 = \pm 169$$

$$\lambda \in \left\{ \frac{16}{8}, \frac{-322}{8} \right\}$$

$$8\left|\sum_{\lambda \in S} \lambda\right| = 8 \times \left| \frac{-306}{8} \right| = 306$$