This purpose of this is show one how to do this type of problem #1

The Euclidean distance, using $u$ as the parameter on the first curve and $v$ as the parameter on the second curve, simplifies to:

$\text{expr}= \sqrt{\left(-\sin (u)-\cos \left(\frac{1}{6} (\pi -6 v)\right)\right)^2+\left(-\cos (u)-\sin \left(\frac{1}{6} (\pi -6 v)\right)-3\right)^2}$

$\frac{\partial \text{expr}}{\partial u} = \frac{\cos \left(u-v+\frac{\pi }{6}\right)-3 \sin (u)}{\sqrt{2 \sin \left(u-v+\frac{\pi }{6}\right)+6 \cos (u)+ \\ 6 \sin \left(\frac{1}{6} (\pi -6 v)\right)+11}}$

$\frac{\partial ^2\text{expr}}{\partial u\, \partial u} = \frac{-40 \sin \left(u-v+\frac{\pi }{6}\right)-6 \sin \left(2 u-v+\frac{\pi }{6}\right)+\sin \left(-2 u+2 v+\frac{\pi }{6}\right) \\ +6 \left(\sin \left(-u+2 v+\frac{\pi }{6}\right)-3 \sin \left(\frac{1}{6} (\pi -6 (u+v))\right)-5\right)-72 \cos (u)-9 \cos (2 u)+9 \sqrt{3} \sin (v)-9 \cos (v)}{2 \left(2 \sin \left(u-v+\frac{\pi }{6}\right)+6 \cos (u)+6 \sin \left(\frac{1}{6} (\pi -6 v)\right)+11\right)^{3/2}}$

$\frac{\partial \text{expr}}{\partial v} = -\frac{\cos \left(u-v+\frac{\pi }{6}\right)+3 \cos \left(\frac{1}{6} (\pi -6 v)\right)}{\sqrt{2 \sin \left(u-v+\frac{\pi }{6}\right)+6 \cos (u)+6 \sin \left(\frac{1}{6} (\pi -6 v)\right)+11}}$

$\frac{\partial ^2\text{expr}}{\partial v\, \partial v} = \frac{-40 \sin \left(u-v+\frac{\pi }{6}\right)-6 \sin \left(2 u-v+\frac{\pi }{6}\right)+\sin \left(-2 u+2 v+ \\ \frac{\pi }{6}\right)+6 \left(\sin \left(-u+2 v+\frac{\pi }{6}\right)-3 \sin \left(\frac{1}{6} (\pi -6 (u+v))\right)-5\right)-18 \cos (u)+9 \sin \left(2 v+\frac{\pi }{6}\right)-72 \sin \left(\frac{1}{6} (\pi -6 v)\right)}{2 \left(2 \sin \left(u-v+\frac{\pi }{6}\right)+6 \cos (u)+6 \sin \left(\frac{1}{6} (\pi -6 v)\right)+11\right)^{3/2}}$

$\frac{\partial ^2\text{expr}}{\partial u\, \partial v} = \frac{\sqrt{3} \sin (2 (u-v))+3 \sqrt{3} \sin (u-2 v)+13 \cos (u-v)-\cos (2 (u-v))+3 \cos (2 u-v)+ \\ 9 \cos (u+v)-3 \cos (u-2 v)-\sqrt{3} (22 \cos (u)+3 \cos (2 u)+9) \sin (v)+2 \sqrt{3} \sin (u) (3 \cos (u)+2) \cos (v)+18 \cos (u)+9 \cos (v)+6}{4 \left(2 \sin \left(u-v+\frac{\pi }{6}\right)+6 \cos (u)+6 \sin \left(\frac{1}{6} (\pi -6 v)\right)+11\right)^{3/2}}$

Depending the needed accuracy and whether you know how to solve such equations symbolically, solve these equations for $u$ and $v$ over $\mathbb{R}$ :

$\text{du}=0\land \text{dv}=0\land 0\leq u<2 \pi \land 0\leq v<2 \pi$

For most such problems, a numerical solution to the equations is sufficient.

The critical points are: $\left.\left\{u\to 0,v\to \frac{2 \pi }{3}\right\},\left\{u\to 0,v\to \frac{5 \pi }{3}\right\},\left\{u\to \pi ,v\to \frac{2 \pi }{3}\right\},\left\{u\to \pi ,v\to \frac{5 \pi }{3}\right\}\right.$ .

$\text{identify}:\,\{\text{duu},\text{dvv},\text{duv}\}\to\text{Block}\left[\left\{D=\text{duu} \text{dvv}-\text{duv}^2\right\}, \\ \text{Which}[D>0\land \text{duu}>0,\text{Minimum},D>0\land \text{duu}<0,\text{Maximum},D<0,\text{Saddle},\text{True},\text{Unknown}]\right]$

$\begin{array}{c} \left\{\left\{0,\frac{2 \pi }{3}\right\},3,\text{Saddle}\right\} \\ \left\{\left\{0,\frac{5 \pi }{3}\right\},5,\text{Maximum}\right\} \\ \left\{\left\{\pi ,\frac{2 \pi }{3}\right\},1,\text{Minimum}\right\} \\ \left\{\left\{\pi ,\frac{5 \pi }{3}\right\},3,\text{Saddle}\right\} \\ \end{array}$

As you may see the minimum is 1. Of course many people do not bother with the second derivative test and sort the extrema by eye. Also, sometimes, higher derivatives are needed .