The centre of mass of a system of two particles divides the distance between them

The centre of mass of a system of two particles of masses $m_1$ and $m_2$ separated by a distance $d$ is:

Center of mass of a system of three particles of masses $1 \, kg, 2 \, kg, 3 \, kg$ is at the point $(1 \, m, 2 \, m, 3 \, m)$ and center of mass of another group of two particles of masses $2 \, kg$ and $3 \, kg$ is at point $(-1 \, m, 3 \, m, -2 \, m)$. Where should a $5 \, kg$ particle be placed so that the center of mass of the system of all these six particles shifts to the center of mass of the first system?

$A$ uniform square plate is shown in the figure. Four identical squares are removed from its corners. If squares $1$ and $3$ are removed,where will the center of mass $(C.M.)$ be located?

Two point masses of $1.5 \ g$ and $2.5 \ g$ are placed at a distance of $16 \ cm$ from each other. If the center of mass is at a distance $x$ from the $1.5 \ g$ mass,then $x = \dots \ cm$.

Define the position vector of the centre of mass.