6. If the three sides $a, b, c$ of $\triangle A B C$ satisfy $a^{2}+b^{2}+3 c^{2}=7$, then the maximum value of the area of $\triangle A B C$ is
Let $D$ be the midpoint of side $AB$ of $\triangle ABC$. By the median length formula, we have $4 CD^{2} + c^{2} = 2(a^{2} + b^{2})$, thus
$$
\begin{array}{l}
2 CD^{2} + \frac{7}{2} c^{2} = a^{2} + b^{2} + 3 c^{2} = 7 \geqslant 2 \sqrt{7} CD \cdot c \\
\geqslant 2 \sqrt{7} c \cdot h_{c} = 4 \sqrt{7} S_{\triangle ABC} \Rightarrow S_{\triangle ABC} \leqslant \frac{\sqrt{7}}{4},
\end{array}
$$

Equality holds when $2 CD^{2} = \frac{7}{2} c^{2} = \frac{7}{2}$ and $a = b$, i.e.,
$c = 1, a = b = \sqrt{2}$. Therefore, the maximum area of $\triangle ABC$ is $\frac{\sqrt{7}}{4}$.