Can all arbuzoids assume the same color?

The problem can be equivalently stated as follows:

Find $a, b, c \in \mathbb N$ such that $$\begin{pmatrix} 13 \\ 17 \\ 15\end{pmatrix} + a\begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} + b\begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix} + c \begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix} \in \left \{ \begin{pmatrix} 45 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 45 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 45 \end{pmatrix} \right \}$$

This amounts to solving three linear systems.

For example, the first one is given by: $$\begin{pmatrix} 13 \\ 17 \\ 15\end{pmatrix} + a\begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} + b\begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix} + c \begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 45 \\ 0 \\ 0 \end{pmatrix}$$ and can be rewritten as $$\begin{cases} 2 a - b - c = 32 \\ -a + 2b - c = -17 \\ -a - b + 2 c = -15 \end{cases}$$ Solving for $a, b, c$ we obtain $$a = n, \qquad b = n - \frac {49} 3, \qquad c = n - \frac {47} 3$$ where $n$ is a parameter. Since $a, b, c$ must be positive integers, there is no solution.

Take the difference between any two colours. For example $G-R=17-13=4$.

Now consider how any colour change will affect this difference. It is unchanged if red+green become blues, will increase by $3$ if blue+red become greens, and will decrease by $3$ if blue+green become reds.

The difference is not itself a multiple of $3$, so it can never become zero by these colour changes. The same is true of any pair of colours, so it isn't even possible for any two colours to become equinumerous, let alone for both to become zero.

The point is that we may consider the free module $M := \mathbb Z^3$. In it, we have the submodule

$$ N := \langle (-1, -1, 2), (-1, 2, -1), (2, -1, -1) \rangle $$

and the residue class

$$ (13,15,17) + N; $$

we want to check whether or not $0$ is in that residue class; this is because we can undo color changes:

$$ (a,b,c) \to (a+2,b-1,c-1) \to (a+1,b-2,c+1) \to (a,b,c), $$

so that $N$ may be assumed a submodule, since it's closed under (additive) inversion. Hence, the problem implies a solution of

$$ (13,15,17) = (2x - y - z, 2y - x - z, 2z - x - y) $$

in integers $x, y, z$. But a clever series of insertions leads to the equation $$ 3(x-y) = 43, $$ which is unsolvable in the integers because $43$ is not divisible by $3$.