jmc

By the Triangle Inequality, $$|z-3i|+|z-4|=|z-4|+|3i-z|\ge |(z-4)+(3i-z)|=|-4+3i|=5.$$But we are told that $|z-3i|+|z-4|=5.$ The only way that equality can occur is if $z$ lies on the line segment connecting 4 and $3i$ in the complex plane.



We want to minimize $|z|$. We see that $|z|$ is minimized when $z$ coincides with the projection of the origin onto the line segment.

The area of the triangle with vertices 0, 4, and $3i$ is $$\frac{1}{2}\cdot 4\cdot 3=6.$$This area is also $$\frac{1}{2}\cdot 5\cdot h=\frac{5h}{2},$$so $h=\boxed{\frac{12}{5}}.$