**Introduction:**
The moment of inertia of a system is a measure of its resistance to rotational motion. In this scenario, we have two point masses, one with mass m and the other with mass 3m, placed at a distance r from each other. We need to determine the moment of inertia of the system about an axis passing through the center of mass (c.o.m) and perpendicular to the line joining the masses.

**Center of Mass (c.o.m):**
The center of mass of a system is the point at which the entire mass of the system can be considered to be concentrated. In this case, since we have two point masses, the c.o.m can be calculated using the formula:

x_com = (m1 * x1 + m2 * x2) / (m1 + m2)

where m1 and m2 are the masses and x1 and x2 are the distances of the masses from the reference point.

**Finding the c.o.m:**
Let's assume that the point mass with mass m is located at a distance x1 from the reference point, and the point mass with mass 3m is located at a distance x2 from the reference point.

Given that the distance between the two masses is r, we can write:

x1 + x2 = r

Since the c.o.m is the reference point, the distances of the masses from the c.o.m are:

x1 = r/2
x2 = r/2

Substituting these values into the formula for the c.o.m, we get:

x_com = (m * r/2 + 3m * r/2) / (m + 3m)
x_com = 2mr / 8m
x_com = r/4

Therefore, the c.o.m of the system is located at a distance of r/4 from the reference point.

**Moment of Inertia about the c.o.m:**
The moment of inertia of a point mass rotating about an axis passing through it is given by the formula:

I = m * r^2

where m is the mass of the point mass and r is the distance of the point mass from the axis of rotation.

In this case, we have two point masses, one with mass m and the other with mass 3m, both located at the same distance of r/4 from the c.o.m. Therefore, the moment of inertia of each point mass about the c.o.m can be calculated as:

I_m1 = m * (r/4)^2
I_m2 = 3m * (r/4)^2

The total moment of inertia of the system about the c.o.m is the sum of the individual moment of inertia values:

I_total = I_m1 + I_m2
I_total = m * (r/4)^2 + 3m * (r/4)^2
I_total = (1/16) * m * r^2 + (9/16) * m * r^2
I_total = (10/16) * m * r^2
I_total = (5/8) * m * r^2

Therefore, the moment of inertia of the system about an axis passing through the c.o.m and perpendicular to the line joining the point masses is (5/8) * m * r^2.

Firstly cal. COM and it is r/4 from 3m mass and 3r/4 from m ... And then cal. IOM = 3/4 Mr^2