Correct Answer - D
`vecv_(CM)=(m_1vecv_1+m_2vecv_2+m_3vecv_3)/(m_1+m_2+m_3)=("Total momentum")/("Total mass")`
Here total momentum of system is zero, beccause momentum of each particle is same is magnitude and they are symmetrically oriented as shown.
`So vecp_1+vecp_2+vecp_3=0`