The problem describes a system in equilibrium with three masses A, B, and C suspended by massless rods. The mass of A is given as 12 kg. We are asked to find the masses of B and C.

1. Problem Description

The problem describes a system in equilibrium with three masses A, B, and C suspended by massless rods. The mass of A is given as 12 kg. We are asked to find the masses of B and C.

2. Solution Steps

Let $m_A$, $m_B$, and $m_C$ be the masses of A, B, and C respectively. We are given $m_A = 12$ kg.

Let $y$ be the distance as shown in the diagram. The length of the horizontal rod connecting A to B and C is $x + 3x = 4x$.

Considering the torques about the pivot point (the top horizontal rod's connection to the supporting vertical rod), we can write the equation for the balance of moments.

$m_A * y = (m_B + m_C) * 3y$

$m_A = 3(m_B + m_C)$

$12 = 3(m_B + m_C)$

$m_B + m_C = 4$

Now, considering the torques about the point where the rod connecting B and C is attached to the rod connecting A, we have:

$m_B * x = m_C * 3x$

$m_B = 3m_C$

Substituting this into the first equation:

$3m_C + m_C = 4$

$4m_C = 4$

$m_C = 1$

Now, substituting the value of $m_C$ back into $m_B = 3m_C$:

$m_B = 3 * 1$

$m_B = 3$

So the masses of B and C are 3 kg and 1 kg respectively.

3. Final Answer

The masses of B and C are 3 kg and 1 kg, respectively. Therefore, the answer is (3) 3 kg, 1 kg.