The perimeter a square whose two sides have equations $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{4}$ and $\frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$ is

Identical direction ratios $2$, $3$, $4$ establish the lines as parallel, implying they are opposite sides of the square.

A point on $L_1$ is $P_1(1, -2, 3)$ and on $L_2$ is $P_2(0, 1, -1)$. The connecting vector is $\vec{P_1P_2} = (-1, 3, -4)$.

The side length $s$ is the perpendicular distance between the parallel lines. Using the direction vector $\vec{d} = (2, 3, 4)$, compute the cross product:

$\vec{P_1P_2} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 3 & -4 \\ 2 & 3 & 4 \end{vmatrix} = 24\mathbf{i} - 4\mathbf{j} - 9\mathbf{k} = (24, -4, -9)$

$|\vec{P_1P_2} \times \vec{d}| = \sqrt{24^2 + (-4)^2 + (-9)^2} = \sqrt{673}$

$|\vec{d}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$

$s = \frac{\sqrt{673}}{\sqrt{29}}$

The perimeter is $4s = 4 \cdot \frac{\sqrt{673}}{\sqrt{29}} = \frac{4 \sqrt{673}}{\sqrt{29}}$, which corresponds to option B.