If the lines x-1 2 = y+1 k = z 2 and x+1 5 = y+1 2 = z k are coplanar, then the equation of the plane containing these lines are

Step-by-step solution

Lines L₁ and L₂ are coplanar if the scalar triple product AB ( d₁ d₂ ) = 0 , where A(1,-1,0) and B(-1,-1,0) are points on each line, and d₁ = (2,k,2) , d₂ = (5,2,k) are their direction vectors. The vector AB = (-2,0,0) , so the determinant condition is: vmatrix -2 & 0 & 0 2 & k & 2 5 & 2 & k vmatrix = 0 Expanding gives -2(k^2 - 4) = 0 , so k^2 = 4 and k = 2 . For k = 2 , the direction vectors are d₁ = (1,1,1) and d₂ = (5,2,2) . The normal vector n = d₁ d₂ = (0,3,-3) (0,1,-1) , yielding the plane equation y - z + 1