Let the plane $$P$$ contain the line $$2x + y - z - 3 = 0 = 5x - 3y + 4z + 9$$ and be parallel to the line $$\frac{x+2}{2} = \frac{3-y}{-4} = \frac{z-7}{5}$$. Then the distance of the point $$A(8, -1, -19)$$ from the plane $$P$$ measured parallel to the line $$\frac{x}{-3} = \frac{y-5}{4} = \frac{z-2}{12}$$ is equal to _____.

Plane $$P \equiv P_1 + \lambda P_2 = 0$$

$$(2x + y - z - 3) + \lambda(5x - 3y + 4z + 9) = 0$$

$$(5\lambda + 2)x + (1 - 3\lambda)y + (4\lambda - 1)z + 9\lambda - 3 = 0$$

Since the normal vector $$\vec{n}$$ is perpendicular to the line with direction vector $$\vec{b} = \langle 2, 4, 5 \rangle$$:

$$\vec{n} \cdot \vec{b} = 0$$

$$2(5\lambda + 2) + 4(1 - 3\lambda) + 5(4\lambda - 1) = 0$$

$$\lambda = -\frac{1}{6}$$

Substituting $$\lambda$$ back gives the equation of Plane P: $$7x + 9y - 10z - 27 = 0$$

Line $$L_1$$ passes through $$A(8, -1, -19)$$ and intersects the plane at point $$B$$

The equation of line AB is: $$\frac{x - 8}{-3} = \frac{y + 1}{4} = \frac{z + 19}{12} = \lambda$$

Let point $$B$$ lie on plane $$P$$. Its parametric coordinates are $$B = (8 - 3\lambda, -1 + 4\lambda, -19 + 12\lambda)$$

Substitute the coordinates of $$B$$ into the equation of plane $$P$$: $$\ 7(8 - 3\lambda) + 9(4\lambda - 1) - 10(12\lambda - 19) = 27$$

$$\lambda = 2$$

$$\therefore \text{Point B} = (2, 7, 5)$$

$$AB = \sqrt{6^2 + 8^2 + 24^2} = 26$$