The direction cosines of the line x-y+2 z=5 and 3 x+y+z=6 are

Step-by-step solution

Direction cosines of the line intersection are determined by the direction vector perpendicular to both plane normals. For planes x - y + 2z = 5 and 3x + y + z = 6 , normal vectors are n₁ = 1, -1, 2 and n₂ = 3, 1, 1 . The direction vector d is their cross product: d = n₁ n₂ = (-1)(1) - (2)(1), (2)(3) - (1)(1), (1)(1) - (-1)(3) = -3, 5, 4 Its magnitude is | d | = (-3)^2 + 5^2 + 4^2 = 50 = 5 2 . The direction cosines are: l = -3 5 2 , m = 5 5 2 , n = 4 5 2 These correspond to option A.