For any three positive real numbers $$a$$, $$b$$ and $$c$$. If $$9(25a^{2} + b^{2}) + 25(c^{2} - 3ac) = 15b(3a + c)$$. Then

We are given that $$9(25a^2 + b^2) + 25(c^2 - 3ac) = 15b(3a + c)$$ for positive reals $$a, b, c$$.

Expanding the left side: $$225a^2 + 9b^2 + 25c^2 - 75ac = 45ab + 15bc$$

Rearranging: $$225a^2 + 9b^2 + 25c^2 - 75ac - 45ab - 15bc = 0$$

We can try to express this as a sum of squares. Notice that:

$$(15a - 3b)^2 = 225a^2 - 90ab + 9b^2$$

$$(3b - 5c)^2 = 9b^2 - 30bc + 25c^2$$

$$(15a - 5c)^2 = 225a^2 - 150ac + 25c^2$$

Adding these three:

$$(15a - 3b)^2 + (3b - 5c)^2 + (15a - 5c)^2 = 450a^2 + 18b^2 + 50c^2 - 90ab - 30bc - 150ac$$

Dividing by 2: $$\frac{1}{2}\left[(15a - 3b)^2 + (3b - 5c)^2 + (15a - 5c)^2\right] = 225a^2 + 9b^2 + 25c^2 - 45ab - 15bc - 75ac$$

This equals our expression, which is 0. Since all terms are squares, each must be zero:

$$15a - 3b = 0 \Rightarrow b = 5a$$

$$3b - 5c = 0 \Rightarrow c = \frac{3b}{5} = 3a$$

$$15a - 5c = 0 \Rightarrow c = 3a$$ (consistent)

So $$a, b, c$$ are in the ratio $$a : 5a : 3a$$, i.e., $$a : b : c = 1 : 5 : 3$$.

Now check which sequence $$b, c, a$$ form: $$b = 5a$$, $$c = 3a$$, $$a = a$$.

For A.P.: $$c - b = 3a - 5a = -2a$$ and $$a - c = a - 3a = -2a$$. Since $$c - b = a - c = -2a$$, the sequence $$b, c, a$$ is in A.P.

Hence, the correct answer is Option B.