1. **Analysis of Assertion A:**
The intensity of light \( I \) can be expressed as:
\[ I = \frac{nh\nu}{A}, \] where \( n \) is the number of photons per unit time, \( h \) is Planck’s constant, \( \nu \) is the frequency, and \( A \) is the area. Rearranging for \( n \):
\[ n = \frac{IA}{h\nu}. \] For a constant intensity \( I \), if the frequency \( \nu \) increases, the number of photons \( n \) decreases. Thus, Assertion A is incorrect.
2. **Analysis of Reason R:**
According to the photoelectric effect, the maximum kinetic energy of emitted electrons is given by:
\[ K_{\text{max}} = h\nu - \phi, \] where \( \phi \) is the work function of the material. As frequency \( \nu \) increases, \( K_{\text{max}} \) also increases. Therefore, Reason R is correct.
Thus, the correct answer is option **(4): Assertion A is not correct, but Reason R is correct.**
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $