6. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with eccentricity $e \in\left[\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{2}}\right]$, the line $y=-x+1$ intersects the ellipse at points $M$ and $N$, and $O$ is the origin with $O M \perp O N$. Then the range of values for $a$ is $\qquad$
6. $\frac{\sqrt{5}}{2} \leqslant a \leqslant \frac{\sqrt{6}}{2}$.

Combining the equations of the line and the ellipse, and simplifying, we get
$$
\begin{array}{l}
\left(a^{2}+b^{2}\right) x^{2}-2 a^{2} x+a^{2}-a^{2} b^{2}=0 . \\
\quad \text { Let } M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right) . \\
\quad \text { Since } O M \perp O N \\
\quad \Rightarrow x_{1} x_{2}+y_{1} y_{2}=0 \\
\quad \Rightarrow 2 x_{1} x_{2}+1-\left(x_{1}+x_{2}\right)=0 .
\end{array}
$$

Using Vieta's formulas and simplifying, we get
$$
\begin{array}{l}
a^{2}+b^{2}=2 a^{2} b^{2} . \\
\text { Also, } \frac{1}{3} \leqslant \frac{c^{2}}{a^{2}} \leqslant \frac{1}{2}, \text { then } \\
\frac{a^{2}}{2} \leqslant b^{2}=\frac{a^{2}}{2 a^{2}-1} \leqslant \frac{2}{3} a^{2} \\
\Rightarrow \frac{\sqrt{5}}{2} \leqslant a \leqslant \frac{\sqrt{6}}{2} .
\end{array}
$$