JEE Main 202427 Jan, Morning shiftMathematicsMediumThree Dimensional Geometry249× since 2002The distance, of the point (7,−2,11)(7,-2,11)(7,−2,11) from the line x−61=y−40=z−83\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}1x−6=0y−4=3z−8 along the line x−52=y−1−3=z−56\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}2x−5=−3y−1=6z−5, is :AOption A. 12BOption B. 18COption C. 21DOption D. 14CheckSkip · show solution
