Let the edges of metal cubes be a₁, a₂ and a₃

And it is given that ratio of edges is 3:4:5

So, let a₁ = 3x, a₂ = 4x and a₃ = 5x

Volume of cube with edge a₁ = v₁ = (a₁)³

⇒v₁ = (3x)³

∴ v₁ = 27x³ cm³→eqn1

Similarly

Volume of cube with edge a₂ = v₂ = (a₂)³

⇒v₂ = (4x)³

∴ v₂ = 64x³ cm³→eqn2

Volume of cube with edge a₃ = v₃ = (a₃)³

⇒v₃ = (5x)³

∴ v₃ = 125x³ cm³→eqn3

Now let the volume of resulting cube be ‘V’ cm³

So, V = v₁ + v₂ + v₃

⇒ V = 27x³ + 64x³ + 125x³ (from eqn1, eqn2 and eqn3)

∴ V = 216x³ cm³→eqn4

It is given that the diagonal of the resulting cube is 12√3 cm

Let the edge of resulting cube be ‘a’ cm

Consider ΔBCD, ∠BDC = 90°

BD = CD = a cm (as they are the edges of cube)

⇒ BC² = a² + a² (putting value of BD and CD)

⇒BC² = 2a²

⇒BC = √(2a²)

∴ BC = a√2 cm

Now consider ΔABC, ∠ABC = 90°

Here, AB = a cm and BC = a√2 cm and AC = 12√3 cm

⇒(12√3)² = a² + (a√2)² (putting values of AB, AC and BC)

⇒ 144 × 3 = a² + 2a²

⇒ 144 × 3 = 3a²

⇒ 144 = a²

⇒ a = √144

∴ a = 12 cm

So, volume of the resulting cube = V = a³

⇒ V = 12³ (putting value of a)

∴ V = 1728 cm³ –eqn5

Equate equation 4 and 5

⇒ 216x³ = 1728

⇒ x³ = 1728/216

⇒ x³ = 8

⇒ x = ∛8

∴ x = 2

So a₁ = 3x = 3 × 2

⇒ a₁ = 6 cm

Similarly a₂ = 4x = 4 × 2

∴ a₂ = 8 cm

Similarly a₃ = 5x = 5 × 2

∴ a₃ = 10 cm

The edges of three cubes are 6 cm, 8 cm and 10 cm.