Let $S=${$z\in \mathrm{ℂ}:\left|z-3\right|\le 1$ and$z\left(4+3i\right)+\overline{z}\left(4-3i\right)\le 24$}. If $\alpha +i\beta$ is the point in $S$ which is closest to $4i$, then $25\left(\alpha +\beta \right)$ is equal to ______.

Given,

$\left|z-3\right|\le 1$

Which represent a circle of radius $1$ and centred at $\left(3,0\right)$

Now simplifying $z\left(4+3i\right)+\overline{z}\left(4-3i\right)\le 24$

We get, $\left(x+iy\right)\left(4+3i\right)+\left(x-iy\right)\left(4-3i\right)\le 24$

$\Rightarrow 4x+3xi+4iy-3y+4x-3ix-4iy-3y\le 24$

$\Rightarrow 8x-6y\le 24$

$\Rightarrow 4x-3y\le 12$

Now plotting the diagram we get,

Minimum distance of $\left(0,4\right)$ from circle $=\sqrt{3^2+4^2}-1=4$ will lie along line joining $\left(0,4\right) \& \left(3,0\right)$
$\therefore$ equation line
$\frac{x}{3}+\frac{y}{4}=1\Rightarrow 4x+3y=12 \dots \left(\mathrm{i}\right)$
Equation of circle ${\left(x-3\right)}^2+y^2=1 \dots \left(\mathrm{ii}\right)$

Solving (i) and (ii) we get,

${\left(\frac{12-3y}{4}-3\right)}^2+y^2=1$

${\left(\frac{-3y}{4}\right)}^2+y^2=1$

$\frac{25y^2}{16}=1\Rightarrow y=\pm \frac{4}{5}$

So, for minimum distance $y=\frac{4}{5}$

$\therefore x=\frac{12}{5}$

$\therefore 25\left(\alpha +\beta \right)=25\left(\frac{4}{5}+\frac{12}{5}\right)$

$\Rightarrow 16\times 5=80$