Problem:
Three rods, each of mass m and length l, are joined together to form an equilateral triangle. We need to find the moment of inertia of the system about an axis passing through its center of mass perpendicular to the plane of the triangle.

Solution:
To find the moment of inertia, we need to consider the mass distribution of the system. Let's break down the solution step by step.

Step 1: Determining the center of mass:
To find the moment of inertia, we first need to determine the center of mass of the system. Since the rods are of equal length and mass, and form an equilateral triangle, the center of mass will be at the intersection point of the medians of the triangle.

Step 2: Identifying the axis of rotation:
The problem states that we need to find the moment of inertia about an axis passing through the center of mass perpendicular to the plane of the triangle. This means that the axis of rotation is passing through the center of mass and is perpendicular to the plane of the equilateral triangle.

Step 3: Finding the moment of inertia of each rod:
To find the moment of inertia of each rod, we can use the formula for the moment of inertia of a rod about its center of mass, which is given by I = (1/12) * m * l^2. Since each rod has a mass of m and a length of l, the moment of inertia of each rod is (1/12) * m * l^2.

Step 4: Applying the parallel-axis theorem:
The parallel-axis theorem states that the moment of inertia of a system about an axis parallel to and a distance d away from an axis through its center of mass is given by I' = I + M * d^2, where I is the moment of inertia about the center of mass, M is the total mass of the system, and d is the perpendicular distance between the two axes.

In our case, we can apply the parallel-axis theorem to find the moment of inertia of the system about the axis passing through its center of mass. Since the three rods are identical, the total mass of the system is 3m. The perpendicular distance between the two axes is the distance from the center of mass to any of the vertices of the equilateral triangle, which is (2/3) * l. Therefore, the moment of inertia of the system about the given axis is I' = 3 * [(1/12) * m * l^2] + 3m * [(2/3) * l]^2.

Step 5: Simplifying the expression:
Simplifying the expression, we get I' = (1/4) * m * l^2 + 2m * l^2.

Final Answer:
Therefore, the moment of inertia of the system about an axis passing through its center of mass perpendicular to the plane of the triangle is given by I' = (1/4) * m * l^2 + 2m * l^2.