JEE Main 202125 Feb, Morning shiftMathematicsMediumThree Dimensional Geometry249× since 2002The equation of the line through the point (0, 1, 2) and perpendicular to the line x−12=y+13=z−1−2{{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 1} \over { - 2}}2x−1=3y+1=−2z−1 is :AOption A. x3=y−1−4=z−23{x \over 3} = {{y - 1} \over { - 4}} = {{z - 2} \over 3}3x=−4y−1=3z−2BOption B. x3=y−14=z−2−3{x \over 3} = {{y - 1} \over 4} = {{z - 2} \over { - 3}}3x=4y−1=−3z−2COption C. x−3=y−14=z−23{x \over { - 3}} = {{y - 1} \over 4} = {{z - 2} \over 3}−3x=4y−1=3z−2DOption D. x3=y−14=z−23{x \over 3} = {{y - 1} \over 4} = {{z - 2} \over 3}3x=4y−1=3z−2CheckSkip · show solution
