Divide $|2z-3(1+i)|=2|z-\frac{3}{2}(1+i)|$, so $S$ is an annulus centred at $C=\frac{3}{2}(1+i)$ with inner radius $\frac{3}{2}$ and outer radius $\frac{7}{2}$.

We seek $\min |z-P|$ where $P=-\frac{1}{2}(5+3i)=\left(-\frac{5}{2},-\frac{3}{2}\right)$.

Distance from $P$ to $C$:

$|PC|=\sqrt{{\left(\frac{3}{2}+\frac{5}{2}\right)}^2+{\left(\frac{3}{2}+\frac{3}{2}\right)}^2}=\sqrt{16+9}=5$

Since $|PC|=5>\frac{7}{2}$ (outer radius), $P$ is outside the annulus. The minimum distance is $|PC|-$ outer radius $=5-\frac{7}{2}=\frac{3}{2}$.