Let the equation of the plane be, A $(x-x_1)$ + B $(y-y_1)$ + C $(z-z_1)$ = 0 Plane passes through the point (-1, 3, 2) ∴ A (x + 1) + B (y - 3) + C (z - 2) = 0 ... (i) Now applying the condition of perpendicularity to the plane (i) with planes x + 2y + 3z = 5 and 3x + 3y + z = 0, we have A + 2B + 3C = 0 3A + 3B + C = 0 Solving we get A + 2B + 3C = 0 9A + 9B + 3C = 0 By cross multiplication, we have, $\frac{A}{2\times 3-9\times 3}$ = $\frac{B}{9\times 3-1\times 3}$ = $\frac{C}{1\times 9-2\times 9}$ ⇒ $\frac{A}{6-27}$ = $\frac{B}{27-3}$ = $\frac{C}{9-18}$ ⇒ $\frac{A}{-21}$ = $\frac{B}{24}$ = $\frac{C}{-9}$ ⇒ $\frac{A}{7}$ = $\frac{B}{-8}$ = $\frac{C}{3}$ ⇒ A = 7λ ; B = - 8λ ; C = 3λ By substituting A and C in equation (i), we get, Substituting the values of A, B and C in equation (i), we have, 7λ (x + 1) - 8λ (y - 3) + 3λ (z - 2) = 0 ⇒ 7x + 7 - 8y + 24 + 3z - 6 = 0 ⇒ 7x - 8y + 3z + 25 = 0