A converging mirror M_1 , a point source S , and a diverging mirror...

For the converging mirror $M_1$, the focal length is $f_1 = -20 \text{ cm}$ and the object distance is $u_1 = -30 \text{ cm}$.

Using the mirror formula for $M_1$:
$\dfrac{1}{v_1} + \dfrac{1}{u_1} = \dfrac{1}{f_1}$
$\dfrac{1}{v_1} - \dfrac{1}{30} = -\dfrac{1}{20} \Rightarrow v_1 = -60 \text{ cm}$

Let the distance between the source $S$ and the diverging mirror $M_2$ be $x$. The distance between the two mirrors is $d = 30 + x$.

For the image to coincide with the source after two reflections (first from $M_1$ and then from $M_2$), the image $I_1$ formed by $M_1$ acts as a virtual object for $M_2$.

The distance of $I_1$ from $M_2$ is $60 - d = 60 - (30 + x) = 30 - x$. Since it is formed behind $M_2$, the object distance for $M_2$ is $u_2 = +(30 - x)$.

The final image is formed at the source $S$, which is at a distance $x$ in front of $M_2$. Thus, the image distance for $M_2$ is $v_2 = -x$.

Using the mirror formula for $M_2$ ($f_2 = +20 \text{ cm}$):
$\dfrac{1}{v_2} + \dfrac{1}{u_2} = \dfrac{1}{f_2}$
$\dfrac{1}{-x} + \dfrac{1}{30 - x} = \dfrac{1}{20}$
$\dfrac{-30 + x + x}{-x(30 - x)} = \dfrac{1}{20}$
$\dfrac{2x - 30}{x^2 - 30x} = -\dfrac{1}{20}$
$-40x + 600 = x^2 - 30x$
$x^2 + 10x - 600 = 0$
$(x + 30)(x - 20) = 0$

Since distance cannot be negative, $x = 20 \text{ cm}$.

To find the location of the image formed by a single reflection from $M_2$, we use $u = -20 \text{ cm}$ and $f = +20 \text{ cm}$:
$\dfrac{1}{v} + \dfrac{1}{-20} = \dfrac{1}{20}$
$\dfrac{1}{v} = \dfrac{1}{20} + \dfrac{1}{20} = \dfrac{1}{10} \Rightarrow v = +10 \text{ cm}$

The positive sign indicates that the image is formed $10 \text{ cm}$ behind the diverging mirror $M_2$. Since $M_1$ is in front of $M_2$, the region behind $M_2$ is farther from the converging mirror.

Answer: $10 \text{ cm}$ from the diverging mirror farther from the converging mirror