Let the minimum value v _ 0 of v = |z| ^ 2 + |z - 3| ^ 2 + |z - 6i|...

Given, the minimum value ${v}_{0}$ of $v = {\left|z\right|}^{2} + {\left|z - 3\right|}^{2} + {\left|z - 6i\right|}^{2}$, $z \in \mathbb{C}$ is attained at $z = {z}_{0}$,

Let $z = x + iy$

$v = {x}^{2} + {y}^{2} + {\left(x - 3\right)}^{2} + {y}^{2} + {x}^{2} + {\left(y - 6\right)}^{2}$

$= \left(3{x}^{2} - 6x + 9\right) + \left(3{y}^{2} - 12y + 36\right)$

$= 3\left({x}^{2} + {y}^{2} - 2x - 4y + 15\right)$

$= 3\left[{\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} + 10\right]$

{V}_\min at $z = 1 + 2i = {z}_{0}$ and ${v}_{0} = 30$

So, ${\left|2{z}_{0}^{2} - \overset{¯}{z}03 + 3\right|}^{2} + {v}_{0}^{2}$

$= {\left|2{\left(1 + 2i\right)}^{2} - {\left(1 - 2i\right)}^{3} + 3\right|}^{2} + 900$

$= {\left|- 6 + 8i - \left(1 + 8i - 6i - 12\right) + 3\right|}^{2} + 900$

$= {\left|8 + 6i\right|}^{2} + 900$ $= 1000$