If the line x+1 1 = y-k 11 = z-4 -5 lies in the plane 2 x+p y+7 z-41=0 which is perpendicular to the plane x+4 y -2 z +13=0 then k =

Step-by-step solution

Given: Line L: x+1 1 = y-k 11 = z-4 -5 with direction vector d = (1, 11, -5) and point P₀(-1, k, 4) . Plane P₁: 2x + py + 7z - 41 = 0 with normal vector n₁ = (2, p, 7) . Plane P₂: x + 4y - 2z + 13 = 0 with normal vector n₂ = (1, 4, -2) . Since P₁ P₂ , their normal vectors are perpendicular: n₁ n₂ = 0 . (2, p, 7) (1, 4, -2) = 2 + 4p - 14 = 0 4p = 12 p = 3 . Substituting p = 3 into P₁ : 2x + 3y + 7z - 41 = 0 . For L to lie in P₁ , point P₀ must satisfy the plane equation: 2(-1) + 3k + 7(4) - 41 = -2 + 3k + 28 - 41 =