Let $ a = (x - y)/3, $ $ b = (y - z)/2, $ and $ c = z $. Then $ x - y = 3a, $ $ y - z = 2b, $ and $ z = c $. Adding these, we get $ x = 3a + 2b + c $. Hence, \[x + \frac{108}{(x - y)^3 (y - z)^2 z} = 3a + 2b + c + \frac{1}{a^3 b^2 c}.\]By AM-GM, \[a + a + a + b + b + c + \frac{1}{a^3 b^2 c} \ge 7.\]Equality occurs when $ a = b = c = 1, $ or $ x = 6, $ $ y = 3, $ and $ z = 1, $ so the minimum value is $ \boxed{7} $.