To find the value of $$f(x) - g(x)$$ at $$x = \frac{3}{2}$$, we need to use the given conditions and properties of differentiable functions.

First, we are told that:

$$f''(x) - g''(x) = 0$$

This implies that:

$$f''(x) = g''(x)$$

Since the second derivatives of both functions are equal, their difference, $$f'(x) - g'(x)$$, must be a linear function. Let’s denote it as:

$$f'(x) - g'(x) = k$$

We'll find the constant $$k$$ using the initial conditions of the derivatives:

$$f'(1) = 2$$

$$g'(1) = 4$$

Thus,

$$f'(1) - g'(1) = 2 - 4 = -2$$

Therefore,

$$f'(x) - g'(x) = -2$$

Integrating the above result, we get:

$$f(x) - g(x) = -2x + C$$

To determine the constant $$C$$, we use the values of the functions at $$x = 2$$:

$$f(2) = 3$$

$$g(2) = 9$$

Thus,

$$f(2) - g(2) = 3 - 9 = -6$$

Therefore,

$$-2 \cdot 2 + C = -6$$

$$-4 + C = -6$$

$$C = -2$$

So the expression for $$f(x) - g(x)$$ is:

$$f(x) - g(x) = -2x - 2$$

We need to find $$f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right)$$:

$$f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right) = -2 \cdot \frac{3}{2} - 2$$

$$= -3 - 2$$

$$= -5$$

Thus, the value of $$f(x) - g(x)$$ at $$x = \frac{3}{2}$$ is -5.