JEE Main 202628 Jan, Evening shiftMathematicsEasyThree Dimensional Geometry249× since 2002Let Q(a,b,c)Q(a, b, c)Q(a,b,c) be the image of the point P(3,2,1)P(3, 2, 1)P(3,2,1) in the line x−11=y2=z−11\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}1x−1=2y=1z−1. Then the distance of Q from the line x−93=y−92=z−5−2\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}3x−9=2y−9=−2z−5 isAOption A. 555BOption B. 888COption C. 666DOption D. 777CheckSkip · show solution