1. Interpret the set $S$ geometrically

Let $z=x+iy$, where $x,y\in\mathbb R$.

The first condition is $|z-3|\le 1.$ Since $3=3+0i$, this represents the closed disk $(x-3)^2+y^2\le 1,$ centered at $(3,0)$ with radius $1$.

2. Simplify the second condition

Given $z(4+3i)+\overline z(4-3i)\le 24.$

Now, $z=x+iy,\qquad \overline z=x-iy.$

Compute: $z(4+3i)=(x+iy)(4+3i)=4x+3ix+4iy+3i^2y=(4x-3y)+i(3x+4y),$

and $\overline z(4-3i)=(x-iy)(4-3i)=4x-3ix-4iy+3i^2y=(4x-3y)-i(3x+4y).$

Adding, $z(4+3i)+\overline z(4-3i)=2(4x-3y)=8x-6y.$

So the inequality becomes $8x-6y\le 24,$ or $4x-3y\le 12.$

Thus $S$ is the part of the disk $(x-3)^2+y^2\le 1$ that lies in the half-plane $4x-3y\le 12.$

3. Point in $S$ closest to $4i$

The point $4i$ is $(0,4)$ in the Argand plane. We must minimize the distance from $(0,4)$ to points in $S$.

First check whether $(0,4)$ lies in the disk: $(0-3)^2+4^2=9+16=25>1,$ so it is outside the disk.

For the full disk, the closest point to $(0,4)$ would lie on the radius from the center $(3,0)$ toward $(0,4)$.

Direction from center to $(0,4)$ is $(-3,4),$ whose magnitude is $5$. Hence the nearest point on the circle is $(3,0)+\frac{1}{5}(-3,4)=\left(3-\frac35,\frac45\right)=\left(\frac{12}{5},\frac45\right).$

Now check whether this point satisfies the half-plane inequality: $4x-3y=4\cdot \frac{12}{5}-3\cdot \frac45=\frac{48}{5}-\frac{12}{5}=\frac{36}{5}=7.2\le 12.$ So this point lies in $S$.

Therefore, the point in $S$ closest to $4i$ is $\alpha+i\beta=\frac{12}{5}+i\frac45.$

Hence, $\alpha=\frac{12}{5},\qquad \beta=\frac45.$

4. Compute $25(\alpha+\beta)$

$\alpha+\beta=\frac{12}{5}+\frac45=\frac{16}{5}.$ Therefore, $25(\alpha+\beta)=25\cdot \frac{16}{5}=5\cdot 16=80.$

5. Comparison with stored answer

Derived answer: $80$. Stored correct answer: $80$.

They match.