If $\left|{\mathrm{z}}_1\right|=2,\left|{\mathrm{z}}_2\right|=3,\left|{\mathrm{z}}_3\right|=4$ and $\left|2{\mathrm{z}}_1+3{\mathrm{z}}_2+4{\mathrm{z}}_3\right|=4$, then $\left|8{\mathrm{z}}_2{\mathrm{z}}_3+27{\mathrm{z}}_3{\mathrm{z}}_1+64{\mathrm{z}}_1{\mathrm{z}}_2\right|=$ 

$\begin{array}{l}\left|8{\mathrm{z}}_2{\mathrm{z}}_3+27{\mathrm{z}}_3{\mathrm{z}}_1+64{\mathrm{z}}_1{\mathrm{z}}_2\right|=\left|{\mathrm{z}}_1\right|\left|{\mathrm{z}}_2\right|\left|{\mathrm{z}}_3\right|\left|\frac{8}{{\mathrm{z}}_1}+\frac{27}{{\mathrm{z}}_2}+\frac{64}{{\mathrm{z}}_3}\right|\\ =\left(2\right)\left(3\right)\left(4\right)\left|\frac{8{\overline{\mathrm{z}}}_1}{{\left|{\mathrm{z}}_1\right|}^2}+\frac{27{\overline{\mathrm{z}}}_2}{{\left|{\mathrm{z}}_2\right|}^2}+\frac{64{\overline{\mathrm{z}}}_3}{{\left|{\mathrm{z}}_3\right|}^2}\right|\\ =24\left|\frac{8{\overline{\mathrm{z}}}_1}{2^2}+\frac{27{\overline{\mathrm{z}}}_2}{3^2}+\frac{64{\overline{\mathrm{z}}}_3}{4^2}\right|\\ \\ =24\left|\overline{2z_1+3z_2+4z_3}\right|\\ \\ =\left(24\right)\left(4\right)=96\end{array}$