=a ‌⇒α+β=2‌‌−2α−2+β=3a ‌⇒β=2−α‌‌−2α−not‌2+not‌2−α=3a⇒α=−a Now both B and C lies as given line ‌α−p⋅2α=21a ‌α(1−2p)=21a ‌−α(1−2p)=21a⇒p=11 ‌β+3+pβ=21a ‌β+3+11β=21a ‌21α+12β+3=0 Also β=2−α ‌21α+12(2−α)+3=0 ‌21α+24−12α+3=0 ‌9α+27=0 ‌α=−3,β=5 So BC=√122 and (BC)2=122