If the image of the point $$P(1, -2, 3)$$ in the plane, $$2x + 3y - 4z + 22 = 0$$ measured parallel to the line, $$\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$$ is $$Q$$, then $$PQ$$ is equal to:

We need to find the image $$Q$$ of the point $$P(1, -2, 3)$$ in the plane $$2x + 3y - 4z + 22 = 0$$, measured parallel to the line $$\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$$.

The direction ratios of the given line are $$(1, 4, 5)$$.

Since the image is measured parallel to this line, we draw a line through $$P$$ with direction ratios $$(1, 4, 5)$$.

The parametric equation of this line through $$P(1, -2, 3)$$ is:

$$x = 1 + t, \quad y = -2 + 4t, \quad z = 3 + 5t$$

This line meets the plane $$2x + 3y - 4z + 22 = 0$$. Substituting:

$$2(1 + t) + 3(-2 + 4t) - 4(3 + 5t) + 22 = 0$$

$$2 + 2t - 6 + 12t - 12 - 20t + 22 = 0$$

$$6 - 6t = 0$$

$$t = 1$$

So the foot of the line on the plane is:

$$F = (1 + 1, \; -2 + 4, \; 3 + 5) = (2, 2, 8)$$

Since $$Q$$ is the image of $$P$$ in the plane (measured along this direction), $$F$$ is the midpoint of $$PQ$$.

$$F = \frac{P + Q}{2}$$

$$Q = 2F - P = (2 \times 2 - 1, \; 2 \times 2 - (-2), \; 2 \times 8 - 3) = (3, 6, 13)$$

Now find $$PQ$$:

$$PQ = \sqrt{(3 - 1)^2 + (6 - (-2))^2 + (13 - 3)^2}$$

$$= \sqrt{4 + 64 + 100} = \sqrt{168} = \sqrt{4 \times 42} = 2\sqrt{42}$$

Therefore, $$PQ = 2\sqrt{42}$$, which is Option B.