10. (20 points) Given the ellipse $\frac{x^{2}}{2}+y^{2}=1$ with its right focus at $F$ and its upper vertex at $M$. Question: Does there exist a line $l$ intersecting the ellipse at points $P$ and $Q$ such that $F$ is the incenter of $\triangle M P Q$? If it exists, find the equation of the line $l$; if not, explain the reason.
10. It is known that $F(1,0)$. Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and the equation of the incircle $\odot F$ of $\triangle M P Q$ is
$$
(x-1)^{2}+y^{2}=r^{2}(r>0) \text {. }
$$

Then the equation of the line $M P$ is
$$
\left(y_{1}-1\right) x-x_{1} y+x_{1}=0 \text {. }
$$

From the fact that $l_{M P}$ is tangent to the circle $\odot F$, we get
$$
\begin{array}{l}
\frac{\left|x_{1}+y_{1}-1\right|}{\sqrt{\left(y_{1}-1\right)^{2}+x_{1}^{2}}}=r \\
\Rightarrow\left(x_{1}+y_{1}-1\right)^{2}=r^{2}\left(x_{1}^{2}+y_{1}^{2}-2 y_{1}+1\right) .
\end{array}
$$

Substituting $x_{1}^{2}+2 y_{1}^{2}=2$ into equation (1), and simplifying, we get
$$
2 x_{1}+\left(r^{2}-1\right) y_{1}+3 r^{2}-3=0 \text {. }
$$

Similarly, we can get that $Q\left(x_{2}, y_{2}\right)$ satisfies
$$
2 x_{2}+\left(r^{2}-1\right) y_{2}+3 r^{2}-3=0 \text {. }
$$

Thus, points $P$ and $Q$ both lie on the line
$$
2 x+\left(r^{2}-1\right) y+3 r^{2}-3=0
$$

i.e., $l_{P Q}: 2 x+\left(r^{2}-1\right) y+3 r^{2}-3=0$.
Since the line $P Q$ is tangent to the circle $\odot F$, we have
$$
\frac{\left|3 r^{2}-1\right|}{\sqrt{\left(r^{2}-1\right)^{2}+4}}=r \Rightarrow r=\sqrt{3}-\sqrt{2} \text {. }
$$

Therefore, the line $P Q$ exists, and its equation is
$$
x+(2-\sqrt{6}) y+6-3 \sqrt{6}=0 .
$$