Step 1: Write the partial fraction identity.
We want to find $ A, B, C $ such that: \[ \frac{1}{(1-2x)^2(1-3x)} = \frac{A}{1-3x} + \frac{B}{1-2x} + \frac{C}{(1-2x)^2} \] Multiply both sides by $ (1-2x)^2(1-3x) $: $ 1 = A(1-2x)^2 + B(1-2x)(1-3x) + C(1-3x) $.
Step 2: Find A by substituting x = 1/3.
When $ x = \frac{1}{3} $, $ 1-3x = 0 $, so the $ B $ and $ C $ terms vanish: $ 1 = A\left(1 - \frac{2}{3}\right)^2 = A \cdot \frac{1}{9} $. Therefore $ A = 9 $.
Step 3: Find C by substituting x = 1/2.
When $ x = \frac{1}{2} $, $ 1-2x = 0 $, so the $ A $ and $ B $ terms vanish: $ 1 = C\left(1 - \frac{3}{2}\right) = C \cdot \left(-\frac{1}{2}\right) $. Therefore $ C = -2 $.
Step 4: Find B by comparing coefficients of x^2.
Expanding the right side, the $ x^2 $ terms are: from $ A(1-2x)^2 $: coefficient $ 4A $; from $ B(1-2x)(1-3x) $: coefficient $ 6B $; from $ C(1-3x) $: no $ x^2 $ term. The left side has no $ x^2 $ term, so $ 4A + 6B = 0 $. With $ A = 9 $: $ 36 + 6B = 0 $, giving $ B = -6 $.
Step 5: List the values.
$ A = 9, B = -6, C = -2 $.
Step 6: Find the minimum and state the answer.
The minimum among $ \{9, -6, -2\} $ is $ -6 $. \[ \boxed{-6} \]