To determine the equation of the line \(PQ\), where \(Q\) is the image of point \(P(2, 3, 4)\) under reflection in the plane given by \(x - 2y + 5z = 6\), we follow these steps:
- Identify the plane equation: \(x - 2y + 5z = 6\). The normal vector to this plane is \(\vec{n} = (1, -2, 5)\).
- Use the reflection formula. The image of a point \(P(x_1, y_1, z_1)\) under reflection in the plane \(Ax + By + Cz + D = 0\) is given by:
\[ Q(x, y, z) = \left( x_1 - \frac{2A(ax_1 + by_1 + cz_1 + d)}{A^2 + B^2 + C^2}, y_1 - \frac{2b(ax_1 + by_1 + cz_1 + d)}{A^2 + B^2 + C^2}, z_1 - \frac{2c(ax_1 + by_1 + cz_1 + d)}{A^2 + B^2 + C^2} \right) \]
- For our plane, \(A = 1\), \(B = -2\), \(C = 5\) and \(D = -6\), and the point \(P(2, 3, 4)\) needs to be considered.
- Plug \(P(2, 3, 4)\) into the plane equation:
\[ 2(1) - 3(2) + 4(5) = 2 - 6 + 20 = 16 \]
- Calculate:
\[ x_{image} = 2 - \frac{2 \cdot 1 \cdot 16}{1^2 + (-2)^2 + 5^2} = 2 - \frac{32}{30} = \frac{28}{15} \] \[ y_{image} = 3 + \frac{2 \cdot 2 \cdot 16}{30} = 3 + \frac{64}{30} = \frac{154}{30} \] \[ z_{image} = 4 - \frac{2 \cdot 5 \cdot 16}{30} = 4 - \frac{160}{30} = \frac{260}{30} \]
- Calculate the coordinates of \(Q\).
- Now determine the direction ratios of line \(PQ\): \[ (x_2 - x_1, y_2 - y_1, z_2 - z_1) = \left(\frac{28}{15} - 2, \frac{154}{30} - 3, \frac{260}{30} - 4\right) \] This simplifies to the direction ratios \((1, -2, 5)\).
- Thus, the line equation in the symmetric form is:
\[ \frac{x-2}{1} = \frac{y-3}{-2} = \frac{z-4}{5} \]
- This matches the correct answer choice: \(\frac{x-2}{1} = \frac{y-3}{-2} = \frac{z-4}{5}\).