Given that the focal length of the concave mirror is $f = 0.15\ \text{m}$. Let the object distance be $u$ and the image distance be $v$. The magnification $m$ is given by $m = \left| \frac{v}{u} \right| = 3$. Therefore, $|v| = 3|u|$, which implies $v = \pm 3u$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$: Case 1: For a real image, $v = -3u$ (where $u$ is the object distance $u_1$): $$ \begin{aligned} \frac{1}{-3u_1} + \frac{1}{u_1} &= \frac{1}{0.15} \\ \Rightarrow \frac{-1 + 3}{3u_1} &= \frac{1}{0.15} \\ \Rightarrow \frac{2}{3u_1} &= \frac{1}{0.15} \\ \Rightarrow 3u_1 &= 0.30 \\ \Rightarrow u_1 &= 0.1\ \text{m} \end{aligned} $$ Case 2: For a virtual image, $v = 3u$ (where $u$ is the object distance $u_2$): $$ \begin{aligned} \frac{1}{3u_2} + \frac{1}{u_2} &= \frac{1}{0.15} \\ \Rightarrow \frac{1 + 3}{3u_2} &= \frac{1}{0.15} \\ \Rightarrow \frac{4}{3u_2} &= \frac{1}{0.15} \\ \Rightarrow 3u_2 &= 0.60 \\ \Rightarrow u_2 &= 0.2\ \text{m} \end{aligned} $$ The distance between the two points is $|u_2 - u_1| = |0.2 - 0.1| = 0.1\ \text{m}$.