Any plane through (–1, 3, 2) isa(x + 1) + b(y – 3) + c (z – 2) = 0 ...(1)Since plane (1) is perpendicular to the planes :x + 2y + 3z = 5 ...(2)and 3x + 3y + z = 0 ...(3)∴ (a)(1) + b(2) + c(3) = 0 and a(3) + b(3) + c(1) = 0⇒ a + 2b + 3c = 0 ...(4)and 3a + 3b + c = 0 ...(5)Solving (4) and (5), we get$\Rightarrow \frac{a}{-7} = \frac{b}{8} = \frac{c}{-3}$Putting in (1), we get –7 (x + 1) + 8 (y – 3) – 3 (z – 2) = 0⇒ –7x –7 + 8y – 24 – 3z + 6 = 0 ⇒ 7x –8y + 3z + 25 = 0,which is the required equation.