6. Given $\alpha, \beta, \gamma \in \mathbf{R}$,
$$
\begin{aligned}
u= & \sin (\alpha-\beta)+\sin (\beta-\gamma)+ \\
& \sin (\gamma-\alpha) .
\end{aligned}
$$

Then $u_{\text {max }}+u_{\text {min }}=$
6.0.
$$
\begin{aligned}
u & =\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \gamma+\sin \gamma \cdot \cos \alpha- \\
& =\left|\begin{array}{lll}
\cos \alpha \cdot \sin \beta-\cos \beta \cdot \sin \gamma-\cos \gamma \cdot \sin \alpha \\
\sin \alpha & \cos \alpha & 1 \\
\sin \beta & \cos \beta & 1 \\
\sin \gamma & \cos \gamma & 1
\end{array}\right| .
\end{aligned}
$$

Let \( A(\sin \alpha, \cos \alpha) \), \( B(\sin \beta, \cos \beta) \), and \( C(\sin \gamma, \cos \gamma) \). Then \( |u| = 2 S_{\triangle ABC} \).
Clearly, \( A \), \( B \), and \( C \) are any three points on the unit circle \( x^2 + y^2 = 1 \). Since the maximum area of \( \triangle ABC \) is \( \frac{3 \sqrt{3}}{4} \) when and only when \( \triangle ABC \) is an equilateral triangle, we have \( |u| \leq \frac{3 \sqrt{3}}{2} \).
Thus, \( u_{\text{max}} = \frac{3 \sqrt{3}}{2} \) and \( u_{\text{min}} = -\frac{3 \sqrt{3}}{2} \). Therefore, \( u_{\text{max}} + u_{\text{min}} = 0 \).