The centre of mass of a system of particles does not depend on

The variation of density of a solid cylindrical rod of cross-sectional area $\alpha$ and length $L$ is given by $\rho = \rho_0 \frac{x^2}{L^2}$, where $x$ is the distance from one end of the rod. The position of its centre of mass from that end $(x=0)$ is:

$A$ slender uniform rod of length $L$ is balanced vertically at a point $P$ on a horizontal surface having some friction. If the top of the rod is displaced slightly to the right,the position of its centre of mass at the time when the rod becomes horizontal is:

On what does the position of the center of mass of a rigid body depend?

Three masses of $2\,kg$,$4\,kg$,and $4\,kg$ are placed at the three points $(1, 0, 0)$,$(1, 1, 0)$,and $(0, 1, 0)$ respectively. The position vector of its center of mass is:

Two particles whose masses are $10\,kg$ and $30\,kg$ and their position vectors are $\hat{i} + \hat{j} + \hat{k}$ and $-\hat{i} - \hat{j} - \hat{k}$ respectively would have the centre of mass at: