The set $|z-5|\le 3$ is the closed disc of radius 3 centred at $(5,0)$. The point on this disc maximising the positive argument is where the tangent from the origin touches the boundary in the upper half-plane.

From the right triangle with hypotenuse 5 (origin to centre) and leg 3 (radius), the perpendicular tangent gives the contact angle $\theta$ at the centre with $\cos \theta =3/5$, $\sin \theta =4/5$ (relative to the centre’s position). So the contact point is $z=(5,0)+3(\cos \theta ,\sin \theta )$—parametrising more carefully:

The contact point lies on the boundary, so $z=5+3e^{i\varphi }$ for some $\varphi$. For the tangent from the origin, the line $Oz$ is perpendicular to the radius, giving the contact direction $\varphi$ such that $z$ has the property below.

Working it out (or using the geometric fact directly): $z=\frac{16}{5}+\frac{12}{5}i$, so $5z-12=16+12i-12=4+12i$ and $5iz+16=5i\left(\frac{16}{5}+\frac{12}{5}i\right)+16=16i-12+16=4+16i$.

${\left|\frac{4+12i}{4+16i}\right|}^2=\frac{16+144}{16+256}=\frac{160}{272}=\frac{10}{17}$.

Multiply by 34: $34\cdot \frac{10}{17}=20$.