JEE Math Practice Question

If the foot of the perpendicular from point (4, 3, 8) on the line $${L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}$$, l $$\ne$$ 0 is (3, 5, 7), then the shortest distance between the line L1 and line $${L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}$$ is equal to :

1. A.$${1 \over {\sqrt 6 }}$$
2. B.$${1 \over 2}$$
3. C.$${1 \over {\sqrt 3 }}$$
4. D.$$\sqrt {{2 \over 3}} $$