The hybridization of orbitals of N atom in $$\text{NO}_3^-$$, $$\text{NO}_2^+$$ and $$\text{NH}_4^+$$ are respectively:

For any central atom, the hybridization can be found from its steric number:
steric number = (number of σ-bonds) + (number of lone pairs).
Hybridization rules: steric number 2 → $$sp$$, 3 → $$sp^2$$, 4 → $$sp^3$$.

Case 1: $$\text{NO}_3^-$$ (nitrate ion)
• Lewis structure: $$[O\!-\!N(=O)\!-\!O]^-$$ with resonance among the three $$N\!-\!O$$ bonds.
• Nitrogen makes 3 σ-bonds with the three oxygens.
• No lone pair remains on nitrogen because its five valence electrons are used in bonding and the positive formal charge on N within the ion’s resonance contributors.
Steric number $$=3$$ ⇒ hybridization $$=sp^2$$.

Case 2: $$\text{NO}_2^+$$ (nitrosonium ion)
• Lewis structure: $$O=N=O^{+}$$ (linear).
• Nitrogen forms 2 σ-bonds (one with each oxygen).
• There is no lone pair on nitrogen; the positive charge indicates loss of one electron.
Steric number $$=2$$ ⇒ hybridization $$=sp$$.

Case 3: $$\text{NH}_4^+$$ (ammonium ion)
• Nitrogen is bonded to four hydrogens through 4 σ-bonds.
• The positive charge shows that nitrogen has donated its lone pair; hence no lone pair is present.
Steric number $$=4$$ ⇒ hybridization $$=sp^3$$.

Thus the hybridizations are: $$sp^2$$ for $$\text{NO}_3^-$$, $$sp$$ for $$\text{NO}_2^+$$ and $$sp^3$$ for $$\text{NH}_4^+$$.

Option A which is: $$sp^2, sp, sp^3$$