Answer: C
The radical (CHX3)X2C∙−CH=CHX2 has a p orbital at C-3 next to the C=C, giving (CHX3)X2C=CH−CHX2∙ as the other resonance form. The unpaired electron is therefore shared between C-3 and C-1 (not C-2), and the two methyl groups on C-3 add six alpha hydrogens for hyperconjugation.