If the roots of z³+iz²+2i = 0 represent the vertices of a $∆$ABC in the Argand plane, then the area of the triangle is

$\begin{array}{l}{z}^3+i{z}^2+2i=0\\ (z-i)\left(z^2+2iz-2\right)=0\\ \Rightarrow z=i, and {\mathrm{z}}^2+2\mathrm{iz}-2=0\\ \begin{array}{l} \\ \Rightarrow (\mathrm{z}+(\mathrm{i}+1\left)\right)(\mathrm{z}+(\mathrm{i}-1\left)\right)=0\\ \Rightarrow \mathrm{z}=-\mathrm{i}-1\text{ and }\mathrm{z}=-\mathrm{i}+1\\ \text{ Hence the triangle is formed by the vertices }\\ \mathrm{A}(0,1),\mathrm{B}(-1,-1),\mathrm{C}(1,-1)\\ \\ Area=\frac{1}{2}\left|\begin{array}{cc}-1-0& 1-0\\ -1-1& -1-1\end{array}\right|=\frac{1}{2}\left(-1(-2)-1(-2)\right)=\frac{4}{2}=2\end{array}\end{array}$