If the foot of the perpendicular from point \( (4, 3, 8) \) on the line \( L_1 : \dfrac{x-a}{1}=\dfrac{y-2}{3}=\dfrac{z-b}{4} \),\(\displaystyle 1 \neq 0\) is \( (3,5,7) \), then the shortest distance between the line \( L_1 \) and line \( L_2 : \dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5} \) is equal to: