<s>[INST] Circumradius of triangle ABC where H is orthocentre and $(AH)(BH)(CH) = 3 $ and $(AH)^2 + (BH)^2 + (CH)^2 = 7 $ 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.
 [/INST] As you wrote, we have
$$AH=2R\cos A,\quad BH=2R\cos B,\quad CH=2R\cos C$$
Now using that
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1\tag1$$
(the proof for $(1)$ is written at the end of this answer)
we get
$$\frac{7}{4R^2}+2\cdot\frac{3}{8R^3}=1,$$
i.e.
$$(R+1)(2R-3)(2R+1)=0$$
to have
$$\color{red}{R=\frac 32}$$

Let us prove $(1)$.
We have
$$\begin{align}\cos 2A+\cos 2B+\cos 2C&=2\cos(A+B)\cos(A-B)+\cos 2C\\\\&=-2\cos C\cos(A-B)+2\cos^2C-1\\\\&=-2\cos C(\cos(A-B)-\cos C)-1\\\\&=-2\cos C\ (\cos(A-B)+\cos(A+B))-1\\\\&=-4\cos A\cos B\cos C-1\end{align}$$
from which we have
$$(2\cos^2A-1)+(2\cos^2B-1)+(2\cos^2C-1)=-4\cos A\cos B\cos C-1$$
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