jmc

Let $L$ be the given mean line. Then, we must have $$\sum _{k=1}^5(z_k-w_k)=0,$$so $$z_1+z_2+z_3+z_4+z_5=w_1+w_2+w_3+w_4+w_5=3+504i.$$Since $L$ has $y$-intercept $3$, it passes through the complex number $3i$, so the points on $L$ can be described parametrically by $3i+zt$, where $z$ is a fixed complex number and $t$ is a real parameter. Let $z_k=3i+z{t}_k$ for each $k$. Then $$z_1+z_2+z_3+z_4+z_5=15i+z(t_1+t_2+t_3+t_4+t_5)=3+504i.$$Setting $t=t_1+t_2+t_3+t_4+t_5$, we have $$zt=3+504i-15i=3+489i,$$so $z=\frac{3}{t}+\frac{489}{t}i$. Thus the slope of $L$ is $\frac{489/t}{3/t}=\boxed{163}$.