Answer: A, B, D
The compound is (2S,3R)-2-bromo-3-methylpentane. Only one conformer places the C-3 hydrogen anti to bromine, and in it the C-2 methyl and the C-3 ethyl groups are cis, giving the Z alkene as the major Saytzeff product, with a little terminal 3-methylpent-1-ene from the C-1 hydrogens. The anti requirement makes the reaction stereospecific, so the (2S,3S) diastereomer gives the E alkene.