We have,

Here mBH×mAC=−1
⇒(α−2β−3)(−21)=−1
⇒β−3=2α−4
⇒β=2α−1 ....(1)
mAH×mBC=−1
⇒(α−1β−2)(−2)=−1
⇒2β−4=α−1
Using (1), we can write:
⇒2(2α−1)−4=α−1
⇒2(2α−1)=α+3
⇒3α=5
α=35,β=37⇒H(35,37)
α+4β=35+328=333=11
β+4α=37+320=327=9
Required equation is
x2−20x+99=0