Let the foot of perpendicular from the point $$(\lambda, 2, 3)$$ on the line $$\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$$ be the point $$(1, \mu, 2)$$. Then the distance between the lines $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$$ and $$\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$$ is equal to :

A

$$\frac{12}{7}$$

B

$$\frac{\sqrt{145}}{7}$$

C

$$\frac{\sqrt{146}}{7}$$

D

$$\frac{\sqrt{143}}{7}$$