Let the foot of perpendicular from the point $$(\lambda, 2, 3)$$ on the line $$\frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$$ be the point $$(1, \mu, 2)$$. Then the distance between the lines $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$$ and $$\frac{x-\lambda}{2} = \frac{y-\mu}{3} = \frac{z+5}{6}$$ is equal to :

To solve this, we first find the constants $$\lambda$$ and $$\mu$$ using the properties of the foot of the perpendicular, and then calculate the distance between the resulting parallel lines.

The foot of the perpendicular from $$P(\lambda, 2, 3)$$ to the line $$L: \frac{x-4}{1} = \frac{y-9}{2} = \frac{z-5}{1}$$ is given as $$Q(1, \mu, 2)$$.

• Step A: Since $$Q(1, \mu, 2)$$ lies on the line $$L$$:
• Step B: The vector $$\vec{PQ} = (1-\lambda, \mu-2, 2-3) = (1-\lambda, 1, -1)$$ must be perpendicular to the line's direction vector $$\vec{v} = (1, 2, 1)$$:
• $$L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$$ (Passes through $$A(1, 2, -4)$$)
• $$L_2: \frac{x-2}{2} = \frac{y-3}{3} = \frac{z+5}{6}$$ (Passes through $$B(2, 3, -5)$$)
• $$\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = \hat{i} + \hat{j} - \hat{k}$$
• $$\vec{AB} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}(6+3) - \hat{j}(6+2) + \hat{k}(3-2) = 9\hat{i} - 8\hat{j} + \hat{k}$$
• $$|\vec{AB} \times \vec{b}| = \sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81 + 64 + 1} = \sqrt{146}$$
• $$|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$

$$\frac{1-4}{1} = \frac{\mu-9}{2} = \frac{2-5}{1}$$

$$-3 = \frac{\mu-9}{2} = -3 \implies \mu - 9 = -6 \implies \mu = 3$$

$$\vec{PQ} \cdot \vec{v} = 0$$

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

$$1 - \lambda + 2 - 1 = 0 \implies \lambda = 2$$

The two lines are:

Both lines have the same direction vector $$\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$$.

The distance $$d$$ between parallel lines is:

$$d = \frac{|\vec{AB} \times \vec{b}|}{|\vec{b}|}$$

$$d = \frac{\sqrt{146}}{7}$$

Correct Option: C