In a triangle ABC with usual notations if, $\cot \frac{A}{2}=\frac{b+c}{a}$, then the triangle $A B C$ is

In a triangle ABC with usual notations, if $3 \mathrm{a}=\mathrm{b}+\mathrm{c}$, then $\cot \frac{\mathrm{B}}{2} \cdot \cot \frac{\mathrm{C}}{2}=$

In $\triangle A B C$, with usual notations, if $\mathrm{a}^4+\mathrm{b}^4+\mathrm{c}^4-2 \mathrm{a}^2 \mathrm{c}^2-2 \mathrm{c}^2 \mathrm{~b}^2=0$, then $\angle \mathrm{C}=\ldots$

In a triangle ABC with usual notations if, $\tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=x \cot \frac{\mathrm{~A}}{2}$, then $x=$