The function $f(x)$ has the property that:

$$f(x) = f(6 - x) \quad \text{for all real } x.$$

We are given the condition:

$$\left(\int_{2}^{3} f(x)\, dx \right)^{2} + \left(\int_{4}^{6} f(x)\, dx \right)^{2} + \left(\int_{2}^{4} f(x)\, dx \right)\left(\int_{0}^{2} f(x)\, dx \right) - \int_{6}^{3} 3f(x)\, dx = -2$$

Solution:

By the symmetry property $f(x) = f(6-x)$, we can simplify the integrals and set $$w = \int_{0}^{3} f(x)\, dx.$$

Then the equation reduces to: $$w^{2} + 3w + 2 = 0.$$

This quadratic factors as: $$(w+2)(w+1) = 0,$$ giving $$w = -1 \quad \text{or} \quad w = -2.$$

The sum of possible values is: $$-1 + (-2) = -3.$$

Answer: (C) $-3$