$A,\mathrm{}B,\mathrm{}C$ lie on the unit circle centered at the origin. Also,
$\frac{3z_1+1.0}{4}=\frac{z_2+z_3}{2}$

i.e. the line segment joining $z_1$ and origin$\left(D\right)$ bisects the line segment joining $z_2 \& z_3$ at $E$

Also, $DE:EA\equiv 3:1$
Let $DE=3K$ and $EA=K$

Now, $DA=4K=1\Rightarrow K=\frac{1}{4}$

$\Rightarrow DE=\frac{3}{4}$
From $△BED,$

$\mathit{\text{BE}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
$\Rightarrow \left|z_2-z_3\right|=BC=\frac{\sqrt{7}}{2}$

Hence, ${\left|z_2-z_3\right|}^2=\frac{7}{4}=1.75$