Let the image of the point $$P(2, -1, 3)$$ in the plane $$x + 2y - z = 0$$ be $$Q$$. Then the distance of the plane $$3x + 2y + z + 29 = 0$$ from the point $$Q$$ is

We need to find the image of point $$P(2, -1, 3)$$ in the plane $$x + 2y - z = 0$$, and then find its distance from another plane.

Recall the formula for the image of a point in a plane.

The image $$Q$$ of a point $$P(x_0, y_0, z_0)$$ in the plane $$ax + by + cz + d = 0$$ is found using the formula:

$$ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} = \frac{-2(ax_0 + by_0 + cz_0 + d)}{a^2 + b^2 + c^2} $$

This formula works because the line from $$P$$ to $$Q$$ is perpendicular to the plane (direction ratios $$(a, b, c)$$), and the midpoint of $$PQ$$ lies on the plane.

Apply the formula with the given values.

Plane: $$x + 2y - z = 0$$, so $$a = 1, b = 2, c = -1, d = 0$$.

Point: $$P(2, -1, 3)$$.

Calculate $$ax_0 + by_0 + cz_0 + d$$:

$$ 1(2) + 2(-1) + (-1)(3) + 0 = 2 - 2 - 3 = -3 $$

Calculate $$a^2 + b^2 + c^2$$:

$$ 1 + 4 + 1 = 6 $$

The parameter value:

$$ t = \frac{-2(-3)}{6} = \frac{6}{6} = 1 $$

Find the coordinates of the image $$Q$$.

$$ x = x_0 + at = 2 + 1(1) = 3 $$

$$ y = y_0 + bt = -1 + 2(1) = 1 $$

$$ z = z_0 + ct = 3 + (-1)(1) = 2 $$

So $$Q = (3, 1, 2)$$.

Verify that $$Q$$ is the correct image.

The midpoint of $$PQ$$ is $$\left(\frac{2+3}{2}, \frac{-1+1}{2}, \frac{3+2}{2}\right) = (2.5, 0, 2.5)$$.

Check if it lies on the plane: $$2.5 + 2(0) - 2.5 = 0$$ . Confirmed.

Find the distance from $$Q(3, 1, 2)$$ to the plane $$3x + 2y + z + 29 = 0$$.

The distance from a point $$(x_0, y_0, z_0)$$ to a plane $$ax + by + cz + d = 0$$ is:

$$ D = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} $$

$$ D = \frac{|3(3) + 2(1) + 1(2) + 29|}{\sqrt{9 + 4 + 1}} = \frac{|9 + 2 + 2 + 29|}{\sqrt{14}} = \frac{42}{\sqrt{14}} $$

Rationalizing:

$$ D = \frac{42}{\sqrt{14}} = \frac{42\sqrt{14}}{14} = 3\sqrt{14} $$

The distance is $$3\sqrt{14}$$.

The correct answer is Option 4: $$3\sqrt{14}$$.