The lines x -3 1 = y -2 1 = z -5 - k and x -4 k = y -3 1 = z -3 2 are coplanar, hence k =

Step-by-step solution

Coplanarity of two lines occurs when the scalar triple product of the connecting vector and the direction vectors is zero. Given lines: x-3 1 = y-2 1 = z-5 -k and x-4 k = y-3 1 = z-3 2 . The determinant condition is: vmatrix x₂-x₁ & y₂-y₁ & z₂-z₁ a₁ & b₁ & c₁ a₂ & b₂ & c₂ vmatrix = 0 With P₁(3,2,5) , d₁ = (1,1,-k) , P₂(4,3,3) , d₂ = (k,1,2) , the connecting vector is (1,1,-2) . Substituting yields: vmatrix 1 & 1 & -2 1 & 1 & -k k & 1 & 2 vmatrix = 0 Expanding the determinant: 1 (1 2 - (-k) 1) - 1 (1 2 - (-k) k) + (