JEE Main 2005Annual, Paper 1MathematicsMediumThree Dimensional Geometry249× since 2002If the angel θ\thetaθ between the line x+11=y−12=z−22{{x + 1} \over 1} = {{y - 1} \over 2} = {{z - 2} \over 2}1x+1=2y−1=2z−2 and the plane 2x−y+λ z+4=02x - y + \sqrt \lambda \,\,z + 4 = 02x−y+λz+4=0 is such that sin θ=13\sin \,\,\theta = {1 \over 3}sinθ=31 then value of λ\lambdaλ is :AOption A. 53{5 \over 3}35BOption B. −35{-3 \over 5}5−3COption C. 34{3 \over 4}43DOption D. −43{-4 \over 3}3−4CheckSkip · show solution
