Given,

2 $$\,\left| \, \right.$$z + 3i$$\,\left| \, \right.$$ = $$\,\left| \, \right.$$z $$-$$i$$\,\left| \, \right.$$

Let z = x + iy

$$ \Rightarrow $$$$\,\,\,$$ 2 $$\,\left| \, \right.$$ x + iy + 3i $$\,\left| \, \right.$$ = $$\,\left| \, \right.$$ x + iy $$-$$ i $$\,\left| \, \right.$$

$$ \Rightarrow $$$$\,\,\,$$ 2 $$\,\left| \, \right.$$ x + i (y + 3)$$\,\left| \, \right.$$ = $$\,\left| \, \right.$$ x + i (y $$-$$ 1)$$\,\left| \, \right.$$

$$ \Rightarrow $$$$\,\,\,$$ 2 $$\sqrt {{x^2} + {{\left( {y + 3} \right)}^2}} $$ = $$\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} $$

$$ \Rightarrow $$$$\,\,\,$$ 4 (x² + y² + 6y + 9) = x² + y² $$-$$ 2y + 1

$$ \Rightarrow $$$$\,\,\,$$ 3x² + 3y² + 26y + 35 = 0

$$ \Rightarrow $$$$\,\,\,$$ x² + y² + $${{26} \over 3}$$ y + $${{35} \over 3}$$ = 0

This is a equation of circle with center ($$-$$ $${{13} \over 3}$$, 0)

$$\therefore\,\,\,$$ Radius = $$\sqrt {0 + {{169} \over 9} - {{35} \over 3}} $$

= $$\sqrt {{{64} \over 9}} $$

= $${8 \over 3}$$