Let the equation of a line through 3 , 6 , - 2 and parallel to the planes x - y + 2 z = 5 and 3 x + y + 2 z = 6 is L = 0 . If point α , β , 2 satisfy L = 0, then α + 2 β is equal to

Step-by-step solution

A normal vector to the 1 s t plane is n 1 → = < 1 , - 1,2 > A normal vector to the 2 n d plane is n 2 → = < 3,1 , 2 > A vector parallel to the required line = i ^ j ^ k ^ 1 - 1 2 3 1 2 = i ^ - 4 - j ^ - 4 + k ^ 4 = - 4 i ^ + 4 j ^ + 4 k ^ Equation of the required line is x - 3 - 1 = y - 6 1 = z + 2 1 ∵ α , β , 2 satisfy the required line ⇒ α - 3 - 1 = β - 6 1 = 4 ⇒ α = - 1 ,   β = 10 ⇒ α + 2 β = 19