Does such a graph exist? [closed]

Question

[EDITED FOR CLARITY]

Does there exist an edge-colored graph $G$ with the following properties?

1. $G$ has a vertex $r$ with exactly three, distinctly colored, incident edges: $(r, u)$, $(r, v)$, $(r, w)$.

2. Every properly colored$^*$ spanning tree $T$ of $G$ has one of the following two types:

   A: $~T$ includes edge $(r, u)$ but neither $(r, v)$ nor $(r, w)$, or

   B: $~T$ includes edges $(r, v)$ and $(r, w)$ but not $(r, u)$.

3. $G$ has at least one properly colored spanning tree of each of these two types.

$^*$(Properly colored means no two incident edges in $T$ have the same color.)

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Here is an example of a graph $G$ that meets every condition except the third (it has just one properly colored connected spanning tree, of type B):

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Here is an example of a graph $G$ where $r$ has two incident edges, and meets the analogous conditions for those two edges (there are properly colored spanning trees with edge $(r, u)$ but not $(r, w)$, or with $(r, w)$ but not $(r, u)$, and every properly colored spanning tree has exactly one of those two edges):

Answer 1

EDIT: The answers below are for previous versions of the question.

Answer for third version:

[This version asked for an edge-colored graph $G$ with a vertex $r$ such that has exactly three edges $a,b,c$, with colors $1,2,3$, and such that $G$ has at least two properly colored spanning trees $T_1$ and $T_2$ such that $T_1$ has $a$ and not $b$ or $c$, and $T_2$ has $b$ and $c$ but not $a$.]

Take $G=(V, E)$ where $V=\{r, u, v, w\}$ and $E=\{(r, u), (r, v), (r, w), (u, v), (v, w)\}$ with edges $(r, u)$, $(v, w)$ colored 1, edges $(r, w)$ and $(u, v)$ colored 3, and edge $(r, v)$ colored 2.

Then take $T_1$ to have edges $(r, v)$, $(r, w)$, $(u, v)$. Take $T_2$ to have edges $(r, u)$, $(u, v)$, $(v, w)$.

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Answer for second version:

[The version asked for an edge-colored graph $G$ with a vertex $r$ with incident edges $a$, $b$, and $c$, each of a different color, that has at least one properly colored spanning tree, and such that every properly colored spanning tree had either edge $a$ (and not $b$ or $c$) or edges $b$ and $c$ (but not $a$).]

Take $G=(V, E)$ where $V=\{r, u, v, w, s\}$ and $E=\{(r, u), (r, v), (r, w), (u, v), (u, s)\}$ with edges $(r, u)$, $(u, s)$ colored 1, edges $(r, w)$ and $(u, v)$ colored 3, and edge $(r, v)$ colored 2. (See the picture below.)

Then the tree $T$ with edges $(r, v)$, $(r, w)$, $(u, s)$, $(u, v)$ form a properly colored spanning tree with edges $b=(r, v)$ and $c=(r, w)$ but not $a=(r, u)$. Also, this is the only properly colored spanning tree.

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Answer for first version:

[This version asked for an edge-colored graph $G$ with a vertex $r$ with incident edges $a$, $b$, and $c$, each of a different color, such that every properly colored spanning tree had either edge $a$ (and not $b$ or $c$) or edges $b$ and $c$ (but not $a$).]

Take $G=(V, E)$ where $V=\{r, u, v, w, s\}$ and $E=\{(r, u), (r, v), (r, w), (u, s)\}$ with edges $(r, u)$, $(u, s)$ colored 1, edge $(r, w)$ colored 3, and edge $(r, v)$ colored 2.

This graph has no properly colored spanning tree, so trivially satisfies the given condition.