Given:

• Masses: $2\,\text{kg}, 4\,\text{kg}, 6\,\text{kg}$
• These are placed at the centres of the three sides of an equilateral triangle.
• Side of triangle: $a=2\,\text{m}$
• Axis: through the centroid of the triangle and perpendicular to its plane.

Key geometry: In an equilateral triangle, the centroid coincides with the incenter. Hence the perpendicular distance from the centroid to each side is the same.

Since the masses are at the centres of the sides, each such point lies on a side, so its distance from the centroid is just the inradius of the triangle.

Find the inradius of the equilateral triangle: For an equilateral triangle of side $a$, $r=\frac{a\sqrt{3}}{6}$ Substituting $a=2$, $r=\frac{2\sqrt{3}}{6}=\frac{\sqrt{3}}{3}\,\text{m}$

Moment of inertia about the given axis: Since all three masses are at the same distance $r$ from the axis, $I=\sum m_i r^2=(2+4+6)r^2$ $I=12\left(\frac{\sqrt{3}}{3}\right)^2$ $I=12\cdot \frac{1}{3}=4\,\text{kg m}^2$

Final answer: $\boxed{4}$

Comparison with stored answer: Stored correct answer = $4$

Our derived answer matches the stored answer.