1. Let us rewrite the given condition in a more useful form.

Given $|z-3-2i|\le 2$ which means $|z-(3+2i)|\le 2.$

So, in the Argand plane, $z$ lies inside or on the circle with:

• center $C=3+2i$
• radius $r=2$

2. Now simplify the expression whose minimum is required:

$|2z-6+5i| = |2(z-3+\tfrac{5}{2}i)|$ But it is cleaner to factor $2$ directly:

$2z-6+5i = 2\left(z-3+\frac{5}{2}i\right)=2\left(z-(3-\tfrac{5}{2}i)\right).$

Hence $|2z-6+5i|=2\left|z-\left(3-\frac{5}{2}i\right)\right|.$

So we need the minimum of the distance from $z$ to the fixed point $P=3-\frac{5}{2}i,$ multiplied by $2$.

3. Find the distance between the center $C=3+2i$ of the allowed circle and the point $P=3-\frac{5}{2}i$.

$CP=\left|\left(3+2i\right)-\left(3-\frac{5}{2}i\right)\right|$ $=\left|2i+\frac{5}{2}i\right|=\left|\frac{9}{2}i\right|=\frac{9}{2}.$

4. The minimum distance from a fixed point outside a circle to any point on or inside the circle is:

$CP-r$ provided $CP>r$.

Here, $CP=\frac{9}{2}, \qquad r=2,$ so $\min \left|z-\left(3-\frac{5}{2}i\right)\right|=\frac{9}{2}-2=\frac{5}{2}.$

5. Therefore,

$\min |2z-6+5i|=2\cdot \frac{5}{2}=5.$

6. Final answer:

$\boxed{5}$

Comparison with stored correct answer:

• Derived answer = $5$
• Stored correct answer = $5$
• They agree.