Compound A reacts with NH$$_4$$Cl and forms a compound B. Compound B reacts with H$$_2$$O and excess of CO$$_2$$ to form compound C which on passing through or reaction with saturated NaCl solution forms sodium hydrogen carbonate. Compound A, B and C, are respectively.

This describes the Solvay process for making sodium hydrogen carbonate (NaHCO₃).

Compound A: Ca(OH)₂ (slaked lime). It reacts with NH₄Cl:

$$Ca(OH)_2 + 2NH_4Cl \to CaCl_2 + 2NH_3 + 2H_2O$$

Compound B: NH₃ (ammonia).

NH₃ reacts with H₂O and excess CO₂:

$$NH_3 + H_2O + CO_2 \to NH_4HCO_3$$

Compound C: NH₄HCO₃ (ammonium hydrogen carbonate).

NH₄HCO₃ reacts with saturated NaCl solution:

$$NH_4HCO_3 + NaCl \to NaHCO_3 + NH_4Cl$$

NaHCO₃ precipitates out (being less soluble).

The correct answer is Option 3: Ca(OH)₂, NH₃, NH₄HCO₃.