Answer: A, B, C
The drawn compound is 3-methylbutan-2-one. Alpha carbons: CH3 (3 H) and CH (1 H), so 4 alpha-hydrogens, not 6 (RDKit count). It is a methyl ketone, so the haloform reaction gives (CH3)2CHCOO(-) and then the acid. Kinetic deprotonation is at the less hindered CH3; the thermodynamic enolate is the more substituted one. So 1, 2, 3 are correct.