Statements :

a) Algebraic sum of moments of mass about centre of mass is equal to zero

b) x-coordinate of centre of mass of system of particles in a plane is represented by ${\mathrm{x}}_{\mathrm{cm}}=\frac{1}{\mathrm{M}}\sum {\mathrm{m}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}$

C) x-coordinate of a rigid body of continuous mass distribution represented by ${\mathrm{x}}_{\mathrm{cm}}=\frac{1}{\mathrm{M}}\int \mathrm{x}.\mathrm{dm}$

The statement "Algebraic sum of moments of mass about the center of mass is equal to zero" is an expression of the principle of conservation of angular momentum. This principle states that, in the absence of external forces acting on a system, the total angular momentum of the system will remain constant. Mathematically, this can be expressed as the sum of all moments of mass about the center of mass being equal to zero.

The formula for the x-coordinate of the center of mass of a system of particles in a plane is given by ${\mathrm{x}}_{\mathrm{cm}}=\frac{1}{\mathrm{M}}\sum {\mathrm{m}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}$, where M is the total mass of the system and $m_i$ and $x_i$ are the mass and x-coordinate of the i_th particle, respectively. 

The formula for the x-coordinate of the center of mass of a rigid body with a continuous mass distribution is given by ${\mathrm{x}}_{\mathrm{cm}}=\frac{1}{\mathrm{M}}\int \mathrm{x}.\mathrm{dm}$, where M is the total mass of the body, x is the position function along the x-axis, and dm represents the infinitesimal mass element at that point. The integral is taken over the entire body. This formula is a generalization of the previous formula for discrete particles.

Hence the correct answer is all a,b,c are true.