JEE Main 202329 Jan, Evening shiftMathematicsMediumThree Dimensional Geometry249× since 2002The shortest distance between the lines x−12=y+8−7=z−45{{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}2x−1=−7y+8=5z−4 and x−12=y−21=z−6−3{{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}2x−1=1y−2=−3z−6 is :AOption A. 232\sqrt323BOption B. 333\sqrt333COption C. 434\sqrt343DOption D. 535\sqrt353CheckSkip · show solution