JEE Main 20179 Apr, Morning shiftMathematicsMediumThree Dimensional Geometry249× since 2002If the line, x−31=y+2−1=z+λ−2{{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}}1x−3=−1y+2=−2z+λ lies in the plane, 2x−4y+3z=2, then the shortest distance between this line and the line, x−112=y9=z4{{x - 1} \over {12}} = {y \over 9} = {z \over 4}12x−1=9y=4z is :AOption A. 2BOption B. 1COption C. 0DOption D. 3CheckSkip · show solution
