JEE Main 2015Session, Shift 1MathematicsMediumThree Dimensional Geometry249× since 2002The distance of the point (1,0,2)(1, 0, 2)(1,0,2) from the point of intersection of the line x−23=y+14=z−212{{x - 2} \over 3} = {{y + 1} \over 4} = {{z - 2} \over {12}}3x−2=4y+1=12z−2 and the plane x−y+z=16,x - y + z = 16,x−y+z=16, is :AOption A. 3213\sqrt {21}321BOption B. 131313COption C. 2142\sqrt {14}214DOption D. 888CheckSkip · show solution
