Let $ P = (a,b,c) $. Then the distance from $ P $ to the plane $ x - z = 0 $ is \[d_1 = \frac{|a - c|}{\sqrt{1^2 + (-1)^2}} = \frac{|a - c|}{\sqrt{2}}.\]The distance from $ P $ to the plane $ x - 2y + z = 0 $ is \[d_2 = \frac{|a - 2b + c|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|a - 2b + c|}{\sqrt{6}}.\]And, the distance from $ P $ to the plane $ x + y + z = 0 $ is \[d_3 = \frac{|a + b + c|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|a + b + c|}{\sqrt{3}}.\]Then the equation $ d_1^2 + d_2^2 + d_3^2 = 36 $ becomes \[\frac{(a - c)^2}{2} + \frac{(a - 2b + c)^2}{6} + \frac{(a + b + c)^2}{3} = 36.\]This simplifies to $ a^2 + b^2 + c^2 = 36 $. Thus, $ S $ is a sphere with radius 6, so its volume is \[\frac{4}{3} \pi \cdot 6^3 = \boxed{288 \pi}.\]