If for the complex numbers   z satisfying | z - 2 - 2 i | ≤ 1 , the maximum value of | 3 i z + 6…

Given that: $|z-2-2i|\le 1$ which represents a circle with centre at $(2,2)$ and radius $=1$.

Now $|3iz+6|=\left|3i\right(z-2i\left)\right|=\left|3i\right||z-2i|$

As we know that, $\left[ \left|z_1{z}_2\right|=\left|z_1\right|\left|z_2\right|\right]$

From tha diagram we can see that, $|3iz+6|=3|z-2i|$ is maximum at $z=3+2i$ because a point on the circle, which has maximum distance from $(0,2)$ is $(3,2)$.

So, $a=3 \& b=2$

$\Rightarrow a+b=5$