Let ABC is an acute angled triangle with orthocentre H. D, E, F are feet of perpendicular from A, B, C on opposite sides. Let R is circumradius of ΔABC. Given $(AH)(BH)(CH) = 3$
and $ (AH)^2 + (BH)^2 + (CH)^2 = 7 $
Then what is the circumradius of triangle?
I know that AH = 2RcosA and so but I get stuck after two or three steps.