Let the line passing through the points $$P(2, -1, 2)$$ and $$Q(5, 3, 4)$$ meet the plane $$x - y + z = 4$$ at the point $$R$$. Then the distance of the point $$R$$ from the plane $$x + 2y + 3z + 2 = 0$$ measured parallel to the line $$\frac{x-7}{2} = \frac{y+3}{2} = \frac{z-2}{1}$$ is _______

Equation of the line passing through $$P$$ and $$Q$$:

$$\frac{x-2}{5-2} = \frac{y-(-1)}{3-(-1)} = \frac{z-2}{4-2} \implies \frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{2} = \lambda$$

Any general point $$R$$ on this line: $$R = (3\lambda + 2, \ 4\lambda - 1, \ 2\lambda + 2)$$

Since $$R$$ lies on the plane $$x - y + z = 4$$: $$(3\lambda + 2) - (4\lambda - 1) + (2\lambda + 2) = 4$$

$$\lambda + 5 = 4 \implies \lambda = -1$$

$$R = (-1, -5, 0)$$

Let $$L$$ be the line passing through $$R(-1, -5, 0)$$ parallel to the line $$\frac{x-7}{2} = \frac{y+3}{2} = \frac{z-2}{1}$$:

$$\frac{x+1}{2} = \frac{y+5}{2} = \frac{z-0}{1} = \mu$$

Any point $$T$$ on this line $$L$$: $$T = (2\mu - 1, \ 2\mu - 5, \ \mu)$$

Since $$T$$ lies on the target plane $$x + 2y + 3z + 2 = 0$$:

$$(2\mu - 1) + 2(2\mu - 5) + 3(\mu) + 2 = 0$$

$$2\mu - 1 + 4\mu - 10 + 3\mu + 2 = 0 \implies 9\mu = 9 \implies \mu = 1$$

$$T = (1, -3, 1)$$

$$RT = \sqrt{(1 - (-1))^2 + (-3 - (-5))^2 + (1 - 0)^2}$$

$$RT = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$$