7. Given $a, b, c \in \mathbf{C}$, and $a+b+c=a^{2}+b^{2}+c^{2}=3, a^{3}+b^{3}+c^{3}=6$, then $(a-1)^{2023}+(b-1)^{2023}+(c-1)^{2023}=$ $\qquad$ .
Let's set $a-1=x, b-1=y, c-1=z$, then $\left\{\begin{array}{l}x+y+z=0, \\ x^{2}+y^{2}+z^{2}=0, \\ x^{3}+y^{3}+z^{3}=3 .\end{array}\right.$ Note that $x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+y z+z x)$,
$$
\begin{array}{l}
x^{3}+y^{3}+z^{3}-3 x y z=(x+y)^{3}-3 x y(x+y)+z^{3}-3 x y z \\
=(x+y+z)\left[(x+y)^{2}+z^{2}-z(x+y)\right]-3 x y(x+y+z) \\
=(x+y+z)\left[x^{2}+y^{2}+z^{2}-(x y+y z+z x)\right],
\end{array}
$$

Thus $\left\{\begin{array}{l}x+y+z=0, \\ x y+y z+z x=0, \Rightarrow x, y, z \text { are the three roots of the cubic equation } x^{3}-1=0 \text {, i.e., }\{x, y, z\}=\left\{1, \omega, \omega^{2}\right\} \text {. } \\ x y z=1\end{array}\right.$ Therefore, $(a-1)^{2023}+(b-1)^{2023}+(c-1)^{2023}=1+\omega^{2023}+\omega^{4046}=1+\omega+\omega^{2}=0$.