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

Given: Point $$P(2, -1, 3)$$, plane $$x + 2y - z = 0$$

Calculating coordinates of image $$Q(x_1, y_1, z_1)$$:

$$\frac{x_1 - 2}{1} = \frac{y_1 - (-1)}{2} = \frac{z_1 - 3}{-1} = -2 \frac{(1)(2) + (2)(-1) + (-1)(3)}{1^2 + 2^2 + (-1)^2}$$

$$\frac{x_1 - 2}{1} = \frac{y_1 + 1}{2} = \frac{z_1 - 3}{-1} = -2 \frac{2 - 2 - 3}{6} = 1$$

$$x_1 - 2 = 1 \implies x_1 = 3$$

$$y_1 + 1 = 2 \implies y_1 = 1$$

$$z_1 - 3 = -1 \implies z_1 = 2$$

Thus, the image point is $$Q(3, 1, 2)$$.

Calculating the perpendicular distance $$d$$ of $$Q(3, 1, 2)$$ from the plane $$3x + 2y + z + 29 = 0$$:

$$d = \frac{\vert{}3(3) + 2(1) + 1(2) + 29\vert{}}{\sqrt{3^2 + 2^2 + 1^2}}$$

$$d = \frac{\vert{}9 + 2 + 2 + 29\vert{}}{\sqrt{14}} = \frac{42}{\sqrt{14}} = 3\sqrt{14}$$