Three identical bar magnets are riveted together at the centre in the same plane as shown in the figure. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet are shown in the figure. Determine the poles of the remaining two.

(N/A) For the system to be in equilibrium,the net force and the net torque acting on the system must be zero.
Let the magnetic moment of each magnet be $M$. The total magnetic moment of the system is the vector sum of the individual magnetic moments: $\vec{M}_{net} = \vec{M}_1 + \vec{M}_2 + \vec{M}_3$.
For the system to be in stable equilibrium in a magnetic field,the net magnetic moment must be zero. If the first magnet is oriented with North at the top,to make the net moment zero,the other two magnets must be arranged such that their components cancel out the moment of the first magnet.
As shown in the solution figure,if magnet $(1)$ has North at the top,then for the other two magnets,the poles must be arranged such that the North poles of magnets $(2)$ and $(3)$ are at the bottom-left and bottom-right respectively,while the South poles are at the top-left and top-right respectively. This configuration ensures that the vector sum of the magnetic moments is zero,resulting in no net torque or force on the system.