The correct decreasing order of energy, for the orbitals having following set of quantum numbers:
(A) $$n = 3, l = 0, m = 0$$
(B) $$n = 4, l = 0, m = 0$$
(C) $$n = 3, l = 1, m = 0$$
(D) $$n = 3, l = 2, m = 1$$

To determine the decreasing order of energy for the given orbitals, we apply the (n + l) rule and first identify each orbital by calculating its (n + l) value. For (A), $$n = 3, l = 0$$ corresponds to the 3s orbital, giving $$n + l = 3$$. For (B), $$n = 4, l = 0$$ corresponds to the 4s orbital, yielding $$n + l = 4$$. For (C), $$n = 3, l = 1$$ corresponds to the 3p orbital, yielding $$n + l = 4$$. Finally, for (D), $$n = 3, l = 2$$ corresponds to the 3d orbital, giving $$n + l = 5$$.

Since higher (n + l) values indicate higher energy, orbital D (with $$n + l = 5$$) has the highest energy, while orbital A (with $$n + l = 3$$) has the lowest. Orbitals B and C both have $$n + l = 4$$, so we compare their n values. Substituting these values shows that B has $$n = 4$$ whereas C has $$n = 3$$, and therefore orbital B lies above orbital C in energy.

From the above, the decreasing order of energy is (D) > (B) > (C) > (A). Thus, the correct answer is Option A: (D) > (B) > (C) > (A).