MathematicsMedium249×since 2002Q2433For real numbers α\alphaα and β\betaβ ≠\ne= 0, if the point of intersection of the straight lines x−α1=y−12=z−13{{x - \alpha } \over 1} = {{y - 1} \over 2} = {{z - 1} \over 3}1x−α=2y−1=3z−1 and x−4β=y−63=z−73{{x - 4} \over \beta } = {{y - 6} \over 3} = {{z - 7} \over 3}βx−4=3y−6=3z−7, lies on the plane x + 2y −-− z = 8, then α\alphaα −-− β\betaβ is equal to :A5B9C3D7Check answerSkip