To find the distance of the plane \(3x + 2y + z + 29 = 0\) from the point \(Q\) which is the image of the point \(P(2, -1, 3)\) in the plane \(x + 2y - z = 0\), we need to follow these steps:
- First, find the image of point \(P\) in the given plane \(x + 2y - z = 0\).
- The formula to find the image of a point \((x_1, y_1, z_1)\) in a plane \(ax + by + cz + d = 0\) is given by: \(x_2 = x_1 - \frac{2a(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}\), \(y_2 = y_1 - \frac{2b(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}\), \(z_2 = z_1 - \frac{2c(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}\).
- For the plane \(x + 2y - z = 0\), we have \(a = 1\), \(b = 2\), \(c = -1\), and \(d = 0\).
- Substitute point \(P(2, -1, 3)\) into the plane equation to find the value: \(1 \cdot 2 + 2 \cdot (-1) - 1 \cdot 3 = 2 - 2 - 3 = -3\).
- Using the formula, we find the coordinates of \(Q(x_2, y_2, z_2)\):
\(x_2 = 2 - \frac{2 \cdot 1 \cdot (-3)}{1^2 + 2^2 + (-1)^2} = 2 + \frac{6}{6} = 3\)
\(y_2 = -1 - \frac{2 \cdot 2 \cdot (-3)}{1^2 + 2^2 + (-1)^2} = -1 + \frac{12}{6} = 1\)
\(z_2 = 3 - \frac{2 \cdot (-1) \cdot (-3)}{1^2 + 2^2 + (-1)^2} = 3 - \frac{6}{6} = 2\)
- Hence, the image \(Q\) is at \((3, 1, 2)\).
- Now, calculate the distance from \(Q(3, 1, 2)\) to the plane \(3x + 2y + z + 29 = 0\).
- The distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(ax + by + cz + d = 0\) is: \(d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\)
- Substitute \((x_1, y_1, z_1) = (3, 1, 2)\), \(a = 3\), \(b = 2\), \(c = 1\), and \(d = 29\) into this formula: \(d = \frac{|3 \cdot 3 + 2 \cdot 1 + 1 \cdot 2 + 29|}{\sqrt{3^2 + 2^2 + 1^2}}\) \(d = \frac{|9 + 2 + 2 + 29|}{\sqrt{14}} = \frac{42}{\sqrt{14}} = 3\sqrt{14}\).
- Therefore, the distance is \(3 \sqrt{14}\), which corresponds to the correct answer option.