If the roots of $${z^3} + i{z^2} + 2i = 0$$    represent the vertices of a $$\vartriangle ABC$$  in the Argand plane then the area of the triangle is

Solution :
$$\eqalign{ & \left( {z - i} \right)\left( {{z^2} + 2iz - 2} \right) = 0 \cr & \Rightarrow \,\,z = i,\frac{{ - 2i \pm \sqrt {4{i^2} + 8} }}{2} = i,1 - i, - 1 - i. \cr} $$
Find the area of the triangle whose vertices are $$\left( {0,1} \right),\left( {1, - 1} \right),\left( { - 1, - 1} \right).$$