To determine the products of the disproportionation of nitrous acid ($$ \mathrm{HNO}_2 $$) at room temperature, let's analyze the chemical behavior of $$ \mathrm{HNO}_2 $$ in aqueous solution.

Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced, yielding two different products. In the case of nitrous acid ($$ \mathrm{HNO}_2 $$), the reaction can be represented as follows:

$$ 3 \mathrm{HNO}_2 \rightarrow \mathrm{HNO}_3 + \mathrm{H_2O} + 2 \mathrm{NO} $$

Breaking this down, we see that nitrous acid disproportionates into nitric acid ($$ \mathrm{HNO}_3 $$), water ($$ \mathrm{H_2O} $$), and nitrogen monoxide ($$ \mathrm{NO} $$). Additionally, in an aqueous solution, nitric acid dissociates to form $$ \mathrm{H_3O}^+ $$ (hydronium ion) and $$ \mathrm{NO_3}^- $$ (nitrate ion).

Therefore, the species formed from this disproportionation reaction are $$ \mathrm{H_3O}^+ $$, $$ \mathrm{NO_3}^- $$, and $$ \mathrm{NO} $$.

From the given options, the correct answer is:

Option A

$$ \mathrm{H_3O}^+ $$, $$ \mathrm{NO_3}^- $$, and $$ \mathrm{NO} $$