The position vectors of two particles of masses $1 \ kg$ and $3 \ kg$ are $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$ respectively. The position vector of the center of mass of this system is:

$A$ uniform square plate $abcd$ has a mass of $1 \, kg$. If two point masses,each of $20 \, g$,are placed at the corners $b$ and $c$ as shown,then the centre of mass shifts on the line

Centre of mass $(C.M.)$ of three particles of masses $1 \,kg, 2 \,kg$ and $3 \,kg$ lies at the point $(1, 2, 3)$ and $C.M.$ of another system of particles of $3 \,kg$ and $2 \,kg$ lies at the point $(-1, 3, -2)$. Where should we put a particle of mass $5 \,kg$ so that the $C.M.$ of the entire system lies at the $C.M.$ of the first system?

$A$ thin uniform wire is bent to form the two equal sides $AB$ and $AC$ of triangle $ABC$,where $AB = AC = 5\,cm$. The third side $BC$,of length $6\,cm$,is made from a uniform wire of twice the linear density of the first. The distance of the centre of mass from $A$ is

$Assertion$ : The position of the centre of mass of a body depends upon the shape and size of the body.
$Reason$ : The centre of mass of a body always lies at the centre of the body.

The center of mass of three particles of masses $10 \ kg, 20 \ kg$,and $30 \ kg$ is at $(0, 0, 0)$. Where should a fourth particle of mass $40 \ kg$ be placed so that the center of mass of the system shifts to $(3, 3, 3)$?