If the roots of the equation ${\mathrm{z}}^3-(2\cos \mathrm{\alpha }+1){\mathrm{z}}^2+(2\cos \mathrm{\alpha }+1)\mathrm{z}-1=0$ are the affixes of the vertices of a triangle in the argand plane (0 < $\mathrm{\alpha }$ < $\mathrm{\pi }$ ) of maximum area then the perimeter of the triangle is

$$\begin{gathered} {\mathrm{z}}^3-(2\cos \mathrm{\alpha }+1){\mathrm{z}}^2+(2\cos \mathrm{\alpha }+1)\mathrm{z}-1=0 \\ \Rightarrow ({\mathrm{z}}^3-1)-(2\cos \mathrm{\alpha }+1)\mathrm{z}(\mathrm{z}-1)=0 \\ \Rightarrow (\mathrm{z}-1)\left({\mathrm{z}}^2+\mathrm{z}+1-\mathrm{z}(2\cos \mathrm{\alpha }+1)\right)=0 \\ \Rightarrow (\mathrm{z}-1)\left({\mathrm{z}}^2-\mathrm{z}(2\cos \mathrm{\alpha })+1)\right) \\ \mathrm{z}=1\text{ is a root then }{\mathrm{z}}^2-(2\cos \mathrm{\alpha })\mathrm{z}+1=0\text{ vertices } \\ \begin{array}{l}(1,0),(\cos \mathrm{\alpha },\sin \mathrm{\alpha }),(\cos \mathrm{\alpha },-\sin \mathrm{\alpha })\\ \mathrm{\Delta }=\text{ Area }=\frac{1}{2}\times 2\sin \mathrm{\alpha }\times (1-\cos \mathrm{\alpha })=\sin \mathrm{\alpha }(1-\cos \mathrm{\alpha })\\ {\mathrm{\Delta }}^{\mathrm{\prime }}=\cos \mathrm{\alpha }-\cos 2\mathrm{\alpha }=0\Rightarrow \mathrm{\alpha }={120}^0\end{array} \end{gathered}$$
$\text{ vertices : }(1,0),\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right),\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$
$\Rightarrow \text{ perimeter }=3\sqrt{3}$