MathematicsMedium249×since 2002Q2497If the foot of the perpendicular from point (4, 3, 8) on the line L1:x−al=y−23=z−b4{L_1}:{{x - a} \over l} = {{y - 2} \over 3} = {{z - b} \over 4}L1:lx−a=3y−2=4z−b, l ≠\ne= 0 is (3, 5, 7), then the shortest distance between the line L₁ and line L2:x−23=y−44=z−55{L_2}:{{x - 2} \over 3} = {{y - 4} \over 4} = {{z - 5} \over 5}L2:3x−2=4y−4=5z−5 is equal to :A16{1 \over {\sqrt 6 }}61B12{1 \over 2}21C13{1 \over {\sqrt 3 }}31D23\sqrt {{2 \over 3}}32Check answerSkip