Given the equation, $ Y = c^\alpha h^\beta G^\gamma $

where Y represents Young's modulus, c the speed of light, h is Planck's constant, G is the gravitational constant, and α, β, and γ are powers to be determined.

We start by expressing the quantities involved in terms of their fundamental dimensions: Mass (M), Length (L), and Time (T). The equation in terms of dimensions is :

$ [M^1 L^{-1} T^{-2}] = [M^0 L^1 T^{-1}]^\alpha [M^1 L^2 T^{-1}]^\beta [M^{-1} L^3 T^{-2}]^\gamma $

Next, by equating the powers of the dimensions M, L, and T on both sides of the equation, we form three separate equations:

1. Equating the powers of M, we get : $ \beta - \gamma = 1 $

2. Equating the powers of L, we get : $ \alpha + 2\beta + 3\gamma = -1 $

3. Equating the powers of T, we get : $ -\alpha - \beta - 2\gamma = -2 $

Now we can solve this system of equations.

From the first equation, we get $ \beta = \gamma + 1 $.

Substituting $ \beta $ from the first equation into the second and third equations, we get :

For the second equation, substituting $ \beta $ we get : $ \alpha + 2(\gamma + 1) + 3\gamma = -1 $, which simplifies to $ \alpha = -5\gamma - 3 $.

For the third equation, substituting $ \beta $ we get : $ -\alpha - (\gamma + 1) - 2\gamma = -2 $, which simplifies to $ \alpha = -3\gamma - 1 $.

Setting the two equations for $ \alpha $ equal to each other gives : $ -5\gamma - 3 = -3\gamma - 1 $, which simplifies to $ \gamma = -2 $.

Substituting $ \gamma = -2 $ back into the equations for $ \alpha $ and $ \beta $ we find $ \alpha = 7 $ and $ \beta = -1 $.

So, the solution to the system is $ \alpha = 7, \beta = -1, \gamma = -2 $, which corresponds to Option A.