JEE Main 202230 Jun, Morning shiftMathematicsHardThree Dimensional Geometry249× since 2002The distance of the point (3, 2, −-−1) from the plane 3x−y+4z+1=03x - y + 4z + 1 = 03x−y+4z+1=0 along the line 2−x2=y−32=z+11{{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}22−x=2y−3=1z+1 is equal to :AOption A. 9BOption B. 6COption C. 3DOption D. 2CheckSkip · show solution
