The equation of the plane containing the line x 1 = y 2 = z 3 and perpendicular to the plane containing the lines x 2 = y 3 = z 1 and x 3 = y 2 = z 1 is

Step-by-step solution

Find the equation of the plane containing line x 1 = y 2 = z 3 and perpendicular to the plane containing lines x 2 = y 3 = z 1 and x 3 = y 2 = z 1 . The direction vector of x 1 = y 2 = z 3 is v_L = i + 2 j + 3 k . Since this line lies in our plane, its direction vector must be perpendicular to the plane's normal vector n₁ = A i + B j + C k , giving the condition: A + 2B + 3C = 0 The direction vectors of the lines in the second plane are v₂ = 2 i + 3 j + k and v₃ = 3 i + 2 j + k . Their cross product gives the norma