The formula for escape velocity is: \[ v_e = \sqrt{\frac{2GM}{R}}, \] where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet. From the formula, it is clear that \( v_e \propto \sqrt{\frac{M}{R}} \). - As the ratio \( \frac{M}{R} \) increases, the escape velocity \( v_e \) increases. Hence, Statement I is correct.
- However, \( v_e \) depends on \( R \) as seen from the formula, so escape velocity is not independent of the radius of the planet.
Hence, Statement II is incorrect. Thus, the correct answer is \( \boxed{(3)} \).
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: