JEE Main 2025April, Unspecified (session-level dataset)MathematicsMediumThree Dimensional Geometry249× since 2002The distance of the point (7,10,11)(7,10,11)(7,10,11) from the line x−41=y−40=z−23\displaystyle\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}1x−4=0y−4=3z−2 along the line x−92=y−13−3=z−176\displaystyle\frac{x-9}{2}=\frac{y-13}{-3}=\frac{z-17}{6}2x−9=−3y−13=6z−17 is:AOption A. 18BOption B. 14COption C. 12DOption D. 16CheckSkip · show solution
