If $\left|{\mathrm{z}}_2+{\mathrm{iz}}_1\right|=\left|{\mathrm{z}}_1\right|+\left|{\mathrm{z}}_2\right|\text{ and }\left|{\mathrm{z}}_1\right|=3$ and $\left|{\mathrm{z}}_2\right|=4$, then the area of $\mathrm{△}\mathrm{ABC}$, if affixes of A, B, and C are z₁, z₂, and [(z₂ - iz₁)/(1 - i)] respectively, is

$\left|{\mathrm{z}}_2+{\mathrm{iz}}_1\right|=\left|{\mathrm{z}}_1\right|+\left|{\mathrm{z}}_2\right|=\left|{\mathrm{iz}}_1\right|+\left|{\mathrm{z}}_2\right|$

$\Rightarrow {\mathrm{iz}}_1,0+\mathrm{i}0\text{ and }{\mathrm{z}}_2\text{ are collinear }$

$\begin{array}{l}\Rightarrow \arg \left({\mathrm{iz}}_1\right)=\arg \left({\mathrm{z}}_2\right)\\ \Rightarrow \arg \left({\mathrm{z}}_2\right)-\arg \left({\mathrm{z}}_1\right)=\frac{\mathrm{\pi }}{2}\end{array}$

Let ${\mathrm{z}}_3=\frac{{\mathrm{z}}_2-{\mathrm{iz}}_1}{1-\mathrm{i}}$

$\begin{array}{l}\mathrm{or} (1-\mathrm{i}){\mathrm{z}}_3={\mathrm{z}}_2-{\mathrm{iz}}_1\\ \mathrm{or} {\mathrm{z}}_2-{\mathrm{z}}_3=\mathrm{i}\left({\mathrm{z}}_1-{\mathrm{z}}_3\right)\end{array}$

$\therefore \mathrm{\angle }\mathrm{ACB}=\frac{\mathrm{\pi }}{2}$

and $\left|{\mathrm{z}}_1-{\mathrm{z}}_3\right|=\left|{\mathrm{z}}_2-{\mathrm{z}}_3\right|$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{AC}=\mathrm{BC}\\ \because \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{AB}}^2={\mathrm{AC}}^2+{\mathrm{BC}}^2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{AC}=\frac{5}{\sqrt{2}} (\because \mathrm{AB}=5)\end{array}$

Therefore, area of $\mathrm{△}\mathrm{ABC}\text{ is }(1/2)\mathrm{AC}\times \mathrm{BC}={\mathrm{AC}}^2/2=25/4\text{ sq. units}$