Step 1: Dimensional analysis.
In dimensional analysis, we express each physical quantity in terms of its fundamental dimensions: mass \( [M] \), length \( [L] \), and time \( [T] \). The dimensions of the variables are:
- \( [G] = \frac{M^{-1}L^3}{T^2} \) (gravitational constant)
- \( [M] = M \) (mass)
- \( [c] = \frac{L}{T} \) (speed of light)
We need to find the dimensions of \( A \). Let the dimensions of \( A \) be \( [A] = L^2 \) (since area is in terms of length squared).
Step 2: Set up the equation.
We are given the equation \( A = G^\alpha M^\beta c^\gamma \), and we need to equate the dimensions of both sides:
\[
[L^2] = \left( \frac{M^{-1}L^3}{T^2} \right)^\alpha \times (M^\beta) \times \left( \frac{L}{T} \right)^\gamma
\]
Now, expanding and simplifying the dimensions:
\[
[L^2] = M^{-\alpha} L^{3\alpha} T^{-2\alpha} \times M^\beta \times L^\gamma T^{-\gamma}
\]
Simplifying the powers of \( M \), \( L \), and \( T \):
\[
[L^2] = M^{-\alpha + \beta} L^{3\alpha + \gamma} T^{-2\alpha - \gamma}
\]
Step 3: Compare powers.
Equating the powers of \( M \), \( L \), and \( T \) on both sides:
- For \( M \), \( -\alpha + \beta = 0 \)
- For \( L \), \( 3\alpha + \gamma = 2 \)
- For \( T \), \( -2\alpha - \gamma = 0 \)
Step 4: Solve the system of equations.
From the equations:
- \( -\alpha + \beta = 0 \Rightarrow \beta = \alpha \)
- \( -2\alpha - \gamma = 0 \Rightarrow \gamma = -2\alpha \)
- \( 3\alpha + \gamma = 2 \Rightarrow 3\alpha - 2\alpha = 2 \Rightarrow \alpha = 2 \)
Thus, \( \alpha = 2 \), \( \beta = 2 \), and \( \gamma = -4 \).
Step 5: Conclusion.
Therefore, the correct answer is (2) \( \alpha = 2, \beta = 2, \gamma = -4 \).