JEE Main 20264 Apr, Morning shiftMathematicsEasyThree Dimensional Geometry249× since 2002The square of the distance of the point (−2,−8,6)(-2,-8,6)(−2,−8,6) from the line x−11=y−12=z−1\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}1x−1=2y−1=−1z along the line x+51=y+5−1=z2\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2}1x+5=−1y+5=2z is equal to:AOption A. 3BOption B. 6COption C. 8DOption D. 12CheckSkip · show solution
