Introduction:
In this problem, we have three point-sized bodies, each with mass M, fixed at the three corners of a light triangular frame with corner length L. We need to find the moment of inertia of these three bodies about an axis perpendicular to the plane of the frame and passing through the center of the frame.

Explanation:
To find the moment of inertia of the three bodies about the given axis, we can use the parallel axis theorem. According to the parallel axis theorem, the moment of inertia about an axis parallel to and at a distance 'd' from an axis passing through the center of mass is given by the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance 'd'.

Step 1: Find the moment of inertia of an individual body about its own axis passing through its center of mass:
The moment of inertia of a point-sized body about its own axis passing through its center of mass is given by the equation:
I = 1/12 * M * R^2
where M is the mass of the body and R is the distance of the body from its axis of rotation.

Step 2: Find the moment of inertia of an individual body about the given axis passing through the center of the frame:
Using the parallel axis theorem, we can calculate the moment of inertia of an individual body about the given axis passing through the center of the frame. Let's assume the distance between the center of the frame and the body is 'd'. Then the moment of inertia of the body about the given axis is given by:
I' = I + M * d^2
where I is the moment of inertia of the body about its own axis passing through its center of mass.

Step 3: Calculate the total moment of inertia of the three bodies:
Since the three bodies are identical, the moment of inertia of each body about the given axis passing through the center of the frame is the same. Therefore, the total moment of inertia of the three bodies is given by:
Total moment of inertia = 3 * I'
where I' is the moment of inertia of an individual body about the given axis passing through the center of the frame.

Conclusion:
In conclusion, the moment of inertia of the three point-sized bodies, each of mass M, fixed at the three corners of a light triangular frame of corner length L, about an axis perpendicular to the plane of the frame and passing through the center of the frame is given by 3 times the moment of inertia of an individual body about the given axis passing through the center of the frame.