On
$t=\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z-3}{4}$ gives $x=1+2t, y=2-3t, z=3+4t$. That is $L_1 = \{ (1,2,3)+t (2,-3,4) \}_t$.
$s=\frac{x+1}{3}=3-y=\frac{z+5}{5}$ gives $x=-1+3s, y=3-s, z=5s-5$. That is $L_2 = \{ (-1,3,-5)+s(3,-1,5)\}_s$.
Hence the distance (squared to avoid roots) between a point on either line is given by $$\|(1,2,3)+t (2,-3,4) - ((-1,3,-5)+s(3,-1,5)) \|^2 = \| ( 2+2t-3s , -1-3t+s , 8+4t-5s ) \|^2$$, which evaluates (assuming I made no mistakes) to $29\,{t}^{2}-58\,s\,t+78\,t+35\,{s}^{2}-94\,s+69$. Differentiating with respect to $s,t$ and setting to zero gives $-58\,t+70\,s-94 = 0$, $58\,t-58\,s+78 = 0 $, and solving yields $s={4 \over 3}, t= -{1 \over 87}$. The corresponding distance is ${16 \sqrt{2} \over \sqrt{87} }$.
An alternative approach could be based on the fact that the vector $(11,-2,-7)$ is perpendicular to the directions of both lines.
As a check I solved for $s,t$ such that the difference between the corresponding points lines up with the vector of the last paragraph, and the $s,t$ values are the same, so any errors are in the first few paragraphs.
Hint : The denominators are direction ratios of the lines. They must be proportional if they are parallel.
I am sure you know the formula for skew lines.