The reaction involves two steps: nucleophilic addition of HCN to form a cyanohydrin, followed by hydrolysis of the cyanide group (-CN) to a carboxylic acid group (-COOH).

1. HCHO (Formaldehyde): $HCHO + HCN \rightarrow H_2C(OH)CN \xrightarrow{H_3O^+} H_2C(OH)COOH$ The product is 2-hydroxyethanoic acid (glycolic acid). The central carbon has two identical hydrogen atoms, so it is achiral and optically inactive.

2. CH3CHO (Acetaldehyde): $CH_3CHO + HCN \rightarrow CH_3CH(OH)CN \xrightarrow{H_3O^+} CH_3CH(OH)COOH$ The product is 2-hydroxypropanoic acid (lactic acid). The central carbon is bonded to four different groups: -H, -OH, -CH3, and -COOH. Thus, it contains an asymmetric carbon (chiral center) and exists as optically active enantiomers.

3. CH3COCH3 (Acetone): $CH_3COCH_3 + HCN \rightarrow (CH_3)_2C(OH)CN \xrightarrow{H_3O^+} (CH_3)_2C(OH)COOH$ The product is 2-hydroxy-2-methylpropanoic acid. The central carbon is bonded to two identical methyl groups, so it is achiral and optically inactive.

Therefore, only acetaldehyde yields an optically active product.