Let the line L pass through (1, 1, 1) and intersect the lines $$\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$$ and $$\frac{x-3}{1} = \frac{y-4}{2} = \frac{z}{1}$$. Then, which of the following points lies on the line L?

Line $$L$$ passes through $$(1, 1, 1)$$ and intersects both lines $$L_1: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$$ and $$L_2: \frac{x-3}{1} = \frac{y-4}{2} = \frac{z}{1}$$.

A general point on $$L_1$$ is $$(1+2t, -1+3t, 1+4t)$$ and on $$L_2$$ is $$(3+s, 4+2s, s)$$.

Let $$L$$ pass through $$(1,1,1)$$ and $$(1+2t, -1+3t, 1+4t)$$ on $$L_1$$. The direction ratios of $$L$$ are $$(2t, 3t-2, 4t)$$.

Parametrically, $$L$$: $$(1+2t\lambda, 1+(3t-2)\lambda, 1+4t\lambda)$$. This must intersect $$L_2$$:

$$1+2t\lambda = 3+s$$ ... (i), $$1+(3t-2)\lambda = 4+2s$$ ... (ii), $$1+4t\lambda = s$$ ... (iii)

From (iii): $$s = 1+4t\lambda$$. Substituting into (i): $$1+2t\lambda = 3+1+4t\lambda = 4+4t\lambda$$

$$-2t\lambda = 3$$, so $$t\lambda = -\frac{3}{2}$$.

From (iii): $$s = 1 + 4\left(-\frac{3}{2}\right) = -5$$.

From (ii): $$1 + (3t-2)\lambda = 4 + 2(-5) = -6$$, so $$(3t-2)\lambda = -7$$.

$$3t\lambda - 2\lambda = -7$$, giving $$3\left(-\frac{3}{2}\right) - 2\lambda = -7$$, so $$-\frac{9}{2} - 2\lambda = -7$$, $$\lambda = \frac{5}{4}$$.

Then $$t = -\frac{3}{2} \cdot \frac{4}{5} = -\frac{6}{5}$$.

Direction of $$L$$: $$(2t, 3t-2, 4t) = \left(-\frac{12}{5}, -\frac{28}{5}, -\frac{24}{5}\right)$$, proportional to $$(3, 7, 6)$$.

Line $$L$$: $$(1+3\mu, 1+7\mu, 1+6\mu)$$. Checking $$(7, 15, 13)$$: $$\mu = 2$$ gives $$(7, 15, 13)$$. ✓

Hence, the correct answer is Option D.