JEE Main 20266 Apr, Evening shiftMathematicsEasyThree Dimensional Geometry249× since 2002The shortest distance between the lines x−41=y−32=z−2−3\dfrac{x-4}{1}=\dfrac{y-3}{2}=\dfrac{z-2}{-3}1x−4=2y−3=−3z−2 and x+22=y−64=z−5−5\dfrac{x+2}{2}=\dfrac{y-6}{4}=\dfrac{z-5}{-5}2x+2=4y−6=−5z−5 is :AOption A. 566\dfrac{5\sqrt6}{6}656BOption B. 252\sqrt525COption C. 353\sqrt535DOption D. 454\sqrt545CheckSkip · show solution
