How hard is it to determine the chromatic number of a unit distance graph?
Okay, this seems easy. Below I sketch why it is NP-hard to decide if a unit distance graph has a $3$-coloring. They key observation is that in a $3$-coloring any two vertices $u$ and $v$ at distance $\sqrt 3$ must get the same color if the two vertices at unit distance from them are also present in the graph. With this gadget we can convert the decision problem of the $3$-colorability of any graph $G$ into the decision problem of the $3$-colorability of a unit distance graph $G'$as follows.

Map the vertices of $G$ into the plane such that no two are at unit distance. Connect every two adjacent vertices with a polygonal path that consists of segments of length $\sqrt 3$ and one segment of length $1$. Add the endpoints of the segments to the vertices of $G'$, and also the two vertices at unit distance from the ends of segments of length $\sqrt 3$. Now $G'$ is $3$-colorable if and only if $G$ is $3$-colorable if there are no additional coincidences among the vertices of $G'$ but we can pick the paths in a generic way.

With a little more care, we can also show that the problem remains NP-complete if we restrict it to planar unit distance graphs. For this, we need to use that it is NP-hard to decide whether a planar graph with maximum degree $4$ has a $3$-coloring. We can repeat the previous construction, just if we map the vertices of $G$ far away in the plane, then the respective paths won't cross. (For this reduction, maximum degree $6$ would suffice.)
This also proves that it's NP-hard to decide whether the chromatic number of a coin graph is $3$ or $4$.

A similar construction also shows that deciding whether a unit distance graph is $4$-colorable is also NP-complete using that there is a unit distance graph in whose only $4$-coloring two given vertices at distance $\frac 83$ must get the same color. Using this gadget, $\frac 83$ can play the role of $\sqrt 3$ in the previous construction.