JEE Main 202313 Apr, Evening shiftMathematicsHardThree Dimensional Geometry249× since 2002The line, that is coplanar to the line x+3−3=y−11=z−55\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}−3x+3=1y−1=5z−5, is :AOption A. x+1−1=y−22=z−54\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}−1x+1=2y−2=4z−5BOption B. x+1−1=y−22=z−55\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}−1x+1=2y−2=5z−5COption C. x−1−1=y−22=z−55\frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}−1x−1=2y−2=5z−5DOption D. x+11=y−22=z−55\frac{x+1}{1}=\frac{y-2}{2}=\frac{z-5}{5}1x+1=2y−2=5z−5CheckSkip · show solution
