JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumThree Dimensional Geometry249× since 2002The perpendicular distance, of the line x−12=y+2−1=z+32\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2}2x−1=−1y+2=2z+3 from the point P(2,−10,1)P(2,-10,1)P(2,−10,1), is:AOption A. 6BOption B. 525\sqrt{2}52COption C. 434\sqrt{3}43DOption D. 353\sqrt{5}35CheckSkip · show solution
