If the system of linear equations $$2x + 2y + 3z = a$$, $$3x - y + 5z = b$$, $$x - 3y + 2z = c$$ where $$a, b, c$$ are non-zero real numbers, has more than one solution, then

For a system of linear equations to have infinitely many solutions, the equations must be linearly dependent. This means we can find non zero constants $$\alpha$$, $$\beta$$, and $$\gamma$$ such that multiplying the three equations by these constants and adding them together results in zero for the coefficients of x, y, and z. Let us set up the relation:

$$\alpha(2x + 2y + 3z) + \beta(3x - y + 5z) + \gamma(x - 3y + 2z) = 0$$

Grouping the terms by the variables x, y, and z, we get:

$$(2\alpha + 3\beta + \gamma)x + (2\alpha - \beta - 3\gamma)y + (3\alpha + 5\beta + 2\gamma)z = 0$$

For this to hold true, the coefficients must individually be equal to zero. This gives us a new system to solve for the multipliers:

$$2\alpha + 3\beta + \gamma = 0,  $$ $$2\alpha - \beta - 3\gamma = 0,  $$ $$3\alpha + 5\beta + 2\gamma = 0.$$

We can solve for the ratio of these constants. Subtracting the second equation from the first gives $$4\beta + 4\gamma = 0$$, which simplifies to $$\beta = -\gamma$$. Substituting $$\beta = -\gamma$$ into the first equation yields $$2\alpha - 3\gamma + \gamma = 0$$, leading to $$2\alpha = 2\gamma$$, or $$\alpha = \gamma$$. Let us assign $$\gamma = 1$$, which immediately makes $$\alpha = 1$$ and $$\beta = -1$$. Checking these values in the third equation gives $$3(1) + 5(-1) + 2(1) = 0$$, which is perfectly consistent.

Since applying these specific weights to the variables on the left side of our original system cancels them out to zero, applying the exact same weights to the constant terms on the right side must also result in zero for the system to be consistent and yield infinitely many solutions. Therefore, we evaluate the relation $$\alpha a + \beta b + \gamma c = 0$$. Substituting our derived values gives $$1a - 1b + 1c = 0$$, which simplifies to $$a - b + c = 0$$. Multiplying the entire equation by negative one to match the given options yields $$b - c - a = 0$$. Therefore, the correct option is B.