Let the foot of perpendicular from the point (λ, 2, 3) on the line...

Since the point $(1, \mu, 2)$ lies on the line $\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}$, we substitute its coordinates into the line equation:

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

$-3 = \dfrac{\mu-9}{2} = -3 \Rightarrow \mu - 9 = -6 \Rightarrow \mu = 3$

The direction ratios of the given line are $(1, 2, 1)$. The direction ratios of the perpendicular from $(\lambda, 2, 3)$ to $(1, 3, 2)$ are $(1-\lambda, 3-2, 2-3) = (1-\lambda, 1, -1)$.

Since the lines are perpendicular, their dot product is zero:

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

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

Substituting $\lambda = 2$ and $\mu = 3$ into the equations of the two lines, we get:

$L_1: \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}$

$L_2: \dfrac{x-2}{2} = \dfrac{y-3}{3} = \dfrac{z+5}{6}$

These are parallel lines with direction vector $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$. The lines pass through the points $\vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}$ and $\vec{a}_2 = 2\hat{i} + 3\hat{j} - 5\hat{k}$ respectively.

The distance $d$ between two parallel lines is given by $d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$.

We have $\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = \hat{i} + \hat{j} - \hat{k}$.

Now, we compute the cross product:

$(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix}$

$= \hat{i}(6 - (-3)) - \hat{j}(6 - (-2)) + \hat{k}(3 - 2) = 9\hat{i} - 8\hat{j} + \hat{k}$

The magnitude of this vector is:

$|9\hat{i} - 8\hat{j} + \hat{k}| = \sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81 + 64 + 1} = \sqrt{146}$

The magnitude of the direction vector $\vec{b}$ is:

$|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

Therefore, the distance between the lines is:

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

Answer: $\dfrac{\sqrt{146}}{7}$