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

$$\begin{gathered} |z-3|\le 1\text{ represent points on or inside the circle of radius }1 \\ \mathrm{\&}\text{ centred at }(3,0) \end{gathered}$$

$\begin{array}{l}z(4+3i)+\overline{z}(4-3i)\le 24\\ (x+iy)(4+3i)+(x-iy)(4-3i)\le 24\\ 4x+3xi+4iy-3y+4x-3ix-4iy-3y\le 24\end{array}$

$\begin{array}{l}8x-6y\le 24\Rightarrow 4x-3y\le 12\\ \end{array}$

$$\begin{gathered} \text{ minimum distance of }(0,4)\text{ from circle }=\sqrt{3^2+4^2}-1=4 \\ \mathrm{will} \mathrm{lie} \mathrm{along} \mathrm{the} \mathrm{line} \mathrm{joining} (0,4)\mathrm{\&}(3,0) \end{gathered}$$

$$\begin{gathered} \frac{x}{3}+\frac{y}{4}=1\Rightarrow 4x+3y=12\dots \\ \text{ equation circle is }(x-3{)}^2+y^2=1\dots \end{gathered}$$

$$\begin{gathered} {\left(\frac{12-3y}{4}-3\right)}^2+y^2=1 \\ \begin{array}{l}{\left(\frac{-3y}{4}\right)}^2+y^2=1\\ \frac{25y^2}{16}=1\Rightarrow y=\pm \frac{4}{5}\\ \text{ for minimum distance }y=\frac{4}{5}\\ \begin{array}{l}\therefore x=\frac{12}{5}\\ \therefore 25(\alpha +\beta )=25\left(\frac{4}{5}+\frac{12}{5}\right)\\ =16\times 5=80\end{array}\end{array} \end{gathered}$$