MathematicsMedium249×since 2002Q2528The shortest distance, between lines L1L_1L1 and L2L_2L2, where L1:x−12=y+1−3=z+42L_1: \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}L1:2x−1=−3y+1=2z+4 and L2L_2L2 is the line, passing through the points A(−4,4,3),B(−1,6,3)\mathrm{A}(-4,4,3), \mathrm{B}(-1,6,3)A(−4,4,3),B(−1,6,3) and perpendicular to the line x−3−2=y3=z−11\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}−2x−3=3y=1z−1, isA141221\frac{141}{\sqrt{221}}221141B24117\frac{24}{\sqrt{117}}11724C42117\frac{42}{\sqrt{117}}11742D121221\frac{121}{\sqrt{221}}221121Check answerSkip