For real numbers α and β ≠ 0 , if the point of intersection of the straight lines x - α 1 = y - 1 2 = z - 1 3 and x - 4 β = y - 6 3 = z - 7 3 lies on the plane x + 2 y - z = 8 , then α - β is equal to :

Step-by-step solution

Given, x - α 1 = y - 1 2 = z - 1 3 and x - 4 β = y - 6 3 = z - 7 3 lies on the plane x + 2 y - z = 8 . Let x - α 1 = y - 1 2 = z - 1 3 = ∅ and x - 4 β = y - 6 3 = z - 7 3 = q Any point on the first line can be considered as ϕ + α ,   2 ϕ + 1 ,   3 ϕ + 1 and a point on the second line can be q β + 4 , 3 q + 6 , 3 q + 7 . For intersection of the lines ϕ + α = q β + 4               . . . 1 2 ϕ + 1 = 3 q