Let L be the line $$\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$$ and let S be the set of all points (a, b, c) on L, whose distance from the line $$\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$$a long the line L is 7. Then $$\sum_{(a,b,c)\in S} (a+b+c) $$ is equal to :

Line $$L$$: $$\frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6} = \lambda$$

Any general point $$P(a, b, c)$$ on $$L$$: $$P = (2\lambda - 1, 3\lambda - 1, 6\lambda - 3)$$

Second Line $$L_2$$: $$\frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0}$$

Let a point on $$L_2$$ be $$Q$$: $$Q = (2\mu - 1, 3\mu - 1, 9)$$

The direction vector of the line segment $$PQ$$ along line $$L$$ is parallel to $$\vec{v}_L = (2, 3, 6)$$:

$$\vec{PQ} = (2\lambda - 2\mu, \ 3\lambda - 3\mu, \ 6\lambda - 12) = k(2, \ 3, \ 6)$$

Comparing components:

$$2(\lambda - \mu) = 2k \implies \lambda - \mu = k$$

$$3(\lambda - \mu) = 3k \implies \lambda - \mu = k$$

$$6\lambda - 12 = 6k \implies \lambda - 2 = k$$

Equating the expressions for $$k$$: $$\lambda - \mu = \lambda - 2 \implies \mu = 2$$

Thus, the point $$Q$$ is: $$Q = (3, 5, 9)$$

Calculating the distance $$PQ = 7$$:

$$PQ = \sqrt{(2\lambda - 1 - 3)^2 + (3\lambda - 1 - 5)^2 + (6\lambda - 3 - 9)^2} = 7$$

$$\sqrt{(2\lambda - 4)^2 + (3\lambda - 6)^2 + (6\lambda - 12)^2} = 7$$

$$\sqrt{4(\lambda - 2)^2 + 9(\lambda - 2)^2 + 36(\lambda - 2)^2} = 7$$

$$\sqrt{49(\lambda - 2)^2} = 7 \implies 7\vert{}\lambda - 2\vert{} = 7 \implies \vert{}\lambda - 2\vert{} = 1$$

$$\lambda - 2 = \pm 1 \implies \lambda_1 = 3, \quad \lambda_2 = 1$$

Finding the points $$P(a,b,c)$$ for both values of $$\lambda$$:

For $$\lambda_1 = 3$$: $$P_1 = (5, 8, 15) \implies a_1 + b_1 + c_1 = 28$$

For $$\lambda_2 = 1$$: $$P_2 = (1, 2, 3) \implies a_2 + b_2 + c_2 = 6$$

$$\sum_{(a,b,c) \in S} (a + b + c) = 28 + 6 = 34$$