<s>[INST] If $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$, then is it unique and its area? Given that $R$ is the circumradius of $\triangle ABC$, and $\cos A=\frac1{2R}$, $\cos B=\frac1{R}$ and $\cos C= \frac3{2R}$. Then would the $\triangle ABC$ be unique? If so how easily we may find its area or otherwise the maximum possible area? 
Any help is appreciated :)
 [/INST] Since
$$
\cos(B)=2\cos(A)\tag{1}
$$
and
$$
\cos(C)=3\cos(A)\tag{2}
$$
and furthermore, since $A+B+C=\pi$,
$$
\cos(C)=-\cos(A+B)\tag{3}
$$
Using $(1)$ and $(2)$ in $(3)$ says
$$
3\cos(A)=\sin(A)\sin(B)-2\cos^2(A)\tag{4}
$$
Moving $2\cos^2(A)$ to the left and squaring yields
$$
\begin{align}
4\cos^4(A)+12\cos^3(A)+9\cos^2(A)&=(1-\cos^2(A))(1-4\cos^2(A))\\
12\cos^3(A)+14\cos^2(A)-1&=0\\
\cos(A)&=0.24312617957188074493\tag{5}
\end{align}
$$
Since $R=\dfrac1{2\cos(A)}$ and the area is $2R^2\sin(A)\sin(B)\sin(C)$, we get
$$
\begin{align}
\text{Area}
&=\frac{\sqrt{1-\cos^2(A)}\sqrt{1-4\cos^2(A)}\sqrt{1-9\cos^2(A)}}{2\cos^2(A)}\\
&=4.90482198456130394784\tag{6}
\end{align}
$$
</s>