Suppose $z = a + bi$ ($a, b \in \mathbb{R}$). By the given condition we can find
$$\begin{aligned} \operatorname{Im} \left( \frac{z-2}{z-i} \right) &= \operatorname{Im} \left( \frac{(a-2)+bi}{a+(b-1)i} \right) \\ &= \frac{-(a-2)(b-1)+ab}{a^2+(b-1)^2} \\ &= \frac{a+2b-2}{a^2+(b-1)^2} = 0, \end{aligned}$$
and thus $a+2b=2$. Therefore,
$$\sqrt{5}|z+3| = \sqrt{(1^2+2^2)((a+3)^2+b^2)} \geq |(a+3)+2b| = 5,$$
namely, $|z+3| \geq \sqrt{5}$. When $a = -2, b = 2$, $|z+3|$ takes the minimum $\sqrt{5}$.

From $\frac{z-2}{z-i} \in \mathbb{R}$ and the geometric meaning of complex division, it is known that the point corresponding to $z$ on the complex plane lies on the line connecting the points corresponding to $2$ and $i$ (excluding the point corresponding to $i$), so the minimum of $|z+3|$ is the distance from point $(-3, 0)$ to line $x + 2y - 2 = 0$ in plane rectangular coordinate system $xOy$, i.e., $\frac{|-3-2|}{\sqrt{1^2+2^2}} = \sqrt{5}$.