Distance Between Parallel Lines Derived from Foot of Perpendicular

To find the distance between the given parallel lines, we first need to determine the values of the constants $\lambda$ and $\mu$.

Step 1: Find the value of $\mu$ The foot of the perpendicular from $P(\lambda, 2, 3)$ to the line $L_1: \frac{x - 4}{1} = \frac{y - 9}{2} = \frac{z - 5}{1}$ is given as $Q(1, \mu, 2)$.

Since $Q(1, \mu, 2)$ lies on $L_1$, its coordinates must satisfy the equation of $L_1$: $\frac{1 - 4}{1} = \frac{\mu - 9}{2} = \frac{2 - 5}{1}$

Simplifying the equal ratios: $-3 = \frac{\mu - 9}{2} = -3$ $\mu - 9 = -6 \implies \mu = 3$

Thus, the point $Q$ is $(1, 3, 2)$.

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Step 2: Find the value of $\lambda$ The vector connecting $P(\lambda, 2, 3)$ and $Q(1, 3, 2)$ is: $\vec{PQ} = (1 - \lambda)\hat{i} + (3 - 2)\hat{j} + (2 - 3)\hat{k} = (1 - \lambda)\hat{i} + \hat{j} - \hat{k}$

Since $PQ$ is perpendicular to line $L_1$, the vector $\vec{PQ}$ is orthogonal to the direction vector of $L_1$, which is $\vec{d}_1 = \hat{i} + 2\hat{j} + \hat{k}$. Therefore, their dot product must be zero: $\vec{PQ} \cdot \vec{d}_1 = 0$ $(1 - \lambda)(1) + (1)(2) + (-1)(1) = 0$ $1 - \lambda + 2 - 1 = 0 \implies \lambda = 2$

Thus, we have $\lambda = 2$ and $\mu = 3$.

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Step 3: Distance between the parallel lines The two parallel lines are: $L_2: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$ $L_3: \frac{x - 2}{2} = \frac{y - 3}{3} = \frac{z + 5}{6}$

Both lines are parallel to the direction vector: $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$

Line $L_2$ passes through the point $A(1, 2, -4)$ and line $L_3$ passes through $B(2, 3, -5)$. The vector $\vec{AB}$ is: $\vec{AB} = (2 - 1)\hat{i} + (3 - 2)\hat{j} + (-5 - (-4))\hat{k} = \hat{i} + \hat{j} - \hat{k}$

The distance $d$ between two parallel lines is given by: $d = \frac{|\vec{AB} \times \vec{b}|}{|\vec{b}|}$

Calculating the cross product $\vec{AB} \times \vec{b}$: $\vec{AB} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix}$ $= \hat{i}(6 - (-3)) - \hat{j}(6 - (-2)) + \hat{k}(3 - 2)$ $= 9\hat{i} - 8\hat{j} + \hat{k}$

Now, calculating the magnitudes: $|\vec{AB} \times \vec{b}| = \sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81 + 64 + 1} = \sqrt{146}$ $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

Therefore, the distance between the lines is: $d = \frac{\sqrt{146}}{7}$

Hence, the correct option is C.