10. Given that the equilateral triangle $\triangle ABC$ is inscribed in the parabola $y=x^{2}-\frac{71}{36}$, and point $P$ satisfies $\overrightarrow{P A}+\overrightarrow{P B}+\overrightarrow{P C}=\overrightarrow{0}$. Does there exist a fixed point $Q$ such that the distance from point $P$ to point $Q$ is equal to the distance from point $P$ to the $x$-axis? If it exists, find the coordinates of point $Q$; if not, explain the reason.
Given $A\left(x_{1}, x_{1}^{2}-\frac{71}{36}\right), B\left(x_{2}, x_{2}^{2}-\frac{71}{36}\right)$,
$C\left(x_{3}, x_{3}^{2}-\frac{71}{36}\right)$, then $k_{A B}=\frac{x_{2}^{2}-x_{1}^{2}}{x_{2}-x_{1}}=x_{1}+x_{2}$,
Similarly, we get $k_{A C}=x_{1}+x_{3}, k_{B C}=x_{2}+x_{3}$.
By the angle formula, we have $\frac{\left(x_{1}+x_{2}\right)-\left(x_{1}+x_{3}\right)}{1+\left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)}=\sqrt{3}$
$$
\Rightarrow x_{2}-x_{3}=\sqrt{3}\left[1+\left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)\right]
$$

Similarly, we get $\frac{\left(x_{2}+x_{3}\right)-\left(x_{1}+x_{2}\right)}{1+\left(x_{1}+x_{2}\right)\left(x_{2}+x_{3}\right)}=\sqrt{3} \Rightarrow x_{3}-x_{1}=\sqrt{3}\left[1+\left(x_{1}+x_{2}\right)\left(x_{2}+x_{3}\right)\right]$;
$$
\frac{\left(x_{1}+x_{3}\right)-\left(x_{2}+x_{3}\right)}{1+\left(x_{1}+x_{3}\right)\left(x_{2}+x_{3}\right)}=\sqrt{3} \Rightarrow x_{1}-x_{2}=\sqrt{3}\left[1+\left(x_{1}+x_{3}\right)\left(x_{2}+x_{3}\right)\right] \text {, }
$$

Then $3+\left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)+\left(x_{1}+x_{2}\right)\left(x_{2}+x_{3}\right)+\left(x_{1}+x_{3}\right)\left(x_{2}+x_{3}\right)=0$
$$
\Rightarrow x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+3\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right)+3=0 \text {. }
$$

Also, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=y_{1}+y_{2}+y_{3}+\frac{71}{12}$, let $P(x, y)$ be the centroid of $\triangle A B C$,
Thus $3\left(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}\right)+6\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right)+6=y_{1}+y_{2}+y_{3}+\frac{71}{12}$
$$
\Rightarrow 3\left(x_{1}+x_{2}+x_{3}\right)^{2}=y_{1}+y_{2}+y_{3}-\frac{1}{12} \Rightarrow 27 x^{2}=3 y-\frac{1}{12} \Rightarrow x^{2}=\frac{1}{9}\left(y-\frac{1}{36}\right) \text {. }
$$

Thus, the trajectory of point $P$ is a parabola, $\frac{p}{2}=\frac{1}{36}$, so its focus is $Q\left(0, \frac{1}{18}\right)$, and the directrix is the $x$-axis. Therefore, there exists a point $Q\left(0, \frac{1}{18}\right)$ that satisfies the condition.