Three balls of masses $$2 \text{ kg}$$, $$4 \text{ kg}$$ and $$6 \text{ kg}$$ respectively are arranged at centre of the edges of an equilateral triangle of side $$2 \text{ m}$$. The moment of inertia of the system about an axis through the centroid and perpendicular to the plane of triangle, will be ___________ $$\text{kgm}^2$$.

We need to find the moment of inertia of three balls placed at the midpoints of the edges of an equilateral triangle, about an axis through the centroid and perpendicular to the plane.

First, recall that the moment of inertia of a system of point masses about an axis is given by $$I = \sum m_i r_i^2$$, where $$r_i$$ is the perpendicular distance of each mass from the axis of rotation.

To determine that distance, note that in an equilateral triangle of side $$a$$, the centroid divides each median in a 2:1 ratio measured from the vertex. Since the median has length $$\frac{\sqrt{3}}{2}a$$, the distance from the centroid to a vertex is $$\frac{2}{3}\times\frac{\sqrt{3}}{2}a = \frac{a}{\sqrt{3}}$$, while the distance from the centroid to the midpoint of a side is $$\frac{1}{3}\times\frac{\sqrt{3}}{2}a = \frac{a}{2\sqrt{3}} = \frac{a\sqrt{3}}{6}$$.

Substituting $$a = 2\text{ m}$$ into the latter expression gives $$r = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}\text{ m}$$.

Because the centroid is equidistant from all three edges, each mass lies at the same distance $$r = \frac{1}{\sqrt{3}}$$ from the axis. Squaring this distance yields $$r^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\,\text{m}^2.$$

Therefore, the total moment of inertia can be written as $$I = m_1r^2 + m_2r^2 + m_3r^2 = (m_1 + m_2 + m_3)\,r^2$$. Substituting the given masses and $$r^2$$ results in $$I = (2 + 4 + 6)\times\frac{1}{3} = 12\times\frac{1}{3} = 4\,\text{kg m}^2.$$

The correct answer is 4 kg m$$^2$$.