JEE Main 202624 Jan, Evening shiftMathematicsEasyThree Dimensional Geometry249× since 2002The sum of all values of α\alphaα, for which the shortest distance between the lines x+1α=y−2−1=z−4−α\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}αx+1=−1y−2=−αz−4 and xα=y−12=z−12α\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha}αx=2y−1=2αz−1 is 2\sqrt22, isAOption A. 8BOption B. -8COption C. -6DOption D. 6CheckSkip · show solution
