6. Given that $\alpha, \beta, \gamma$ are the three interior angles of a triangle, if $\cos \alpha+\cos \beta+\cos \gamma$ attains its maximum value, then $\sin \alpha+\sin \beta+$ $\sin \gamma=$ $\qquad$
$$
\begin{array}{l}
\cos \alpha + \cos \beta + \cos \gamma = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} - 2 \cos^2 \frac{\alpha + \beta}{2} + 1 \\
\leqslant -2 \cos^2 \frac{\alpha + \beta}{2} + 2 \cos \frac{\alpha + \beta}{2} + 1 = -2 \left( \cos \frac{\alpha + \beta}{2} - \frac{1}{2} \right)^2 + \frac{3}{2} \leqslant \frac{3}{2},
\end{array}
$$

Equality holds when $\alpha = \beta$ and $\cos \frac{\alpha + \beta}{2} = \frac{1}{2} \Rightarrow \alpha = \beta = \gamma = 60^{\circ}$. 
Therefore, in this case, $\sin \alpha + \sin \beta + \sin \gamma = \frac{3 \sqrt{3}}{2}$.