Let the line $$L$$ pass through the point (0, 1, 2), intersect the line $$\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}$$ and be parallel to the plane $$2x + y - 3z = 4$$. Then the distance of the point $$P(1, -9, 2)$$ from the line $$L$$ is

The line $$L$$ passes through $$(0, 1, 2)$$, intersects the line $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$, and is parallel to the plane $$2x + y - 3z = 4$$.

Find the intersection point.

A general point on the given line is $$(1 + 2\lambda, 2 + 3\lambda, 3 + 4\lambda)$$.

Direction ratios of $$L$$ from $$(0, 1, 2)$$ to this point: $$(1 + 2\lambda,\ 1 + 3\lambda,\ 1 + 4\lambda)$$.

Apply the parallel-to-plane condition.

Since $$L$$ is parallel to the plane $$2x + y - 3z = 4$$, its direction is perpendicular to the normal $$(2, 1, -3)$$:

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

$$2 + 4\lambda + 1 + 3\lambda - 3 - 12\lambda = 0 \implies -5\lambda = 0 \implies \lambda = 0$$

Determine line $$L$$.

The intersection point is $$(1, 2, 3)$$, so $$L$$ passes through $$(0, 1, 2)$$ and $$(1, 2, 3)$$ with direction $$(1, 1, 1)$$.

Distance from $$P(1, -9, 2)$$ to line $$L$$.

$$\vec{AP} = P - A = (1 - 0,\ -9 - 1,\ 2 - 2) = (1, -10, 0)$$

$$\vec{d} = (1, 1, 1)$$

$$\vec{AP} \times \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -10 & 0 \\ 1 & 1 & 1 \end{vmatrix} = (-10 - 0)\vec{i} - (1 - 0)\vec{j} + (1 + 10)\vec{k} = (-10, -1, 11)$$

$$|\vec{AP} \times \vec{d}|^2 = 100 + 1 + 121 = 222$$

$$|\vec{d}|^2 = 1 + 1 + 1 = 3$$

$$\text{Distance} = \frac{\sqrt{222}}{\sqrt{3}} = \sqrt{\frac{222}{3}} = \sqrt{74}$$

The answer is Option A: $$\sqrt{74}$$.