If $\left|3z-2\right|=5$ then the locus of $\mathrm{z}$ is

Given $\left|3z-2\right|=5$

Substitute $z=x+iy$

so that $\left|3x-2+i\left(3y\right)\right|=5$

Use the modulus formula of complex number 

$\sqrt{{\left(3x-2\right)}^2+9y^2}=5$

Squaring on both sides 

$9x^2+9y^2-12x+4=25$

Therefore, the locus of the given complex number is $\boxed{9x^2+9y^2-12x-21=0}$