If the area of a triangle with vertices Z₁, Z₂ and Z₃ is the absolute value of the number $\mathrm{\lambda i}\left|\begin{array}{l}{\mathrm{Z}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\overline{\mathrm{Z}}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{Z}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\overline{\mathrm{Z}}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{Z}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\overline{\mathrm{Z}}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|$ then the value of $1/\mathrm{\lambda }$ is equal to ______.

Let ${\mathrm{z}}_1={\mathrm{x}}_1+{\mathrm{iy}}_1,{\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2\text{ and }{\mathrm{z}}_3={\mathrm{x}}_3+{\mathrm{iy}}_3$ be the vertices.
Consider $\left|\begin{array}{ccc}{\mathrm{z}}_1& {\overline{\mathrm{z}}}_1& 1\\ {\mathrm{z}}_2& {\overline{\mathrm{z}}}_2& 1\\ {\mathrm{z}}_3& {\overline{\mathrm{z}}}_3& 1\end{array}\right|=\left|\begin{array}{l}{\mathrm{x}}_1+{\mathrm{iy}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_1-{\mathrm{iy}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{x}}_2+{\mathrm{iy}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_2-{\mathrm{iy}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{x}}_3+{\mathrm{iy}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_3-{\mathrm{iy}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|{\mathrm{C}}_1\to {\mathrm{C}}_1+{\mathrm{C}}_2$

$=2\left|\begin{array}{l}{\mathrm{x}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_1-{\mathrm{iy}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{x}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_2-{\mathrm{iy}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{x}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_3-{\mathrm{iy}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1$
$\begin{array}{l}=\left|\begin{array}{lll}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|=(-4\mathrm{i})\frac{1}{2}\left|\right|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\mid \\ =-4\mathrm{i}\times \text{ Area of triangle. }\end{array}$
$\therefore$Area of triangle $=\frac{\mathrm{i}}{4}\left|\begin{array}{l}{\mathrm{z}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\overline{\mathrm{z}}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{z}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\overline{\mathrm{z}}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ {\mathrm{z}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\overline{\mathrm{z}}}_3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right|;\therefore \mathrm{\lambda }=\frac{1}{4}\Rightarrow \frac{1}{\mathrm{\lambda }}=4$