Explanation

(c) To keep the centre of mass at the position, velocity of centre of mass is zero, so
$\frac{m_{1}\mathbf{v}_{\mathbf{1}} + m_{2}\mathbf{v}_{\mathbf{2}}}{m_{1} + m_{2}} = 0$
where $\mathbf{v}_{\mathbf{1}}\ and\ \mathbf{v}_{\mathbf{2}}$ are velocities of particles 1 and 2 respectively.
$\Rightarrow$ $m_{1}\frac{d\mathbf{r}_{\mathbf{1}}}{dt} + m_{2}\frac{d\mathbf{r}_{2}}{dt} = 0$
$\lbrack\because\mathbf{v}_{\mathbf{1}} = \frac{d\mathbf{r}_{\mathbf{1}}}{dt}\ and\ \mathbf{v}_{\mathbf{2}} = \frac{d\mathbf{r}_{\mathbf{2}}}{dt}\rbrack$
$\Rightarrow$ $md\mathbf{r}_{\mathbf{1}} + m_{2}d\mathbf{r}_{\mathbf{2}} = 0\ \lbrack d\mathbf{r}_{\mathbf{1}}\ and\ d\mathbf{r}_{2}$ represent the change in displacement of particles]
Let 2nd particle has been displaced by distance $x$.
$\Rightarrow$ $m_{1}(d) + m_{2}(x) = 0$
$\Rightarrow$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - \frac{m_{1}d}{m_{2}}$
Negative sign shows that both the particles have to move in opposite directions.
So, $\frac{m_{1}d}{m_{2}}$ is the distance moved by 2nd particle to keep centre of mass at the same position.