10. (20 points) Given a fixed point $M(-3,0)$, and $Q$, $P$ are moving points on the $x$-axis and $y$-axis respectively, such that $M P \perp P Q$. Point $N$ is on the line $P Q$, and $\overrightarrow{P N}=-\frac{3}{2} \overrightarrow{N Q}$.
(1) Find the equation of the trajectory $C$ of the moving point $N$.
(2) Draw a line $l$ through the point $T(-1,0)$ that intersects the trajectory $C$ at two points $A$ and $B$. Is there a point $D$ on the $x$-axis such that $\triangle A B D$ is an equilateral triangle? If it exists, find the coordinates of point $D$; if not, explain the reason.
10. (1) Let points $N(x, y)$, $P\left(0, y^{\prime}\right)$, and $Q\left(x^{\prime}, 0\right)$ with $\left(x^{\prime}>0\right)$.
From $\overrightarrow{P N}=-\frac{3}{2} \overrightarrow{N Q}$, we get $x^{\prime}=\frac{x}{3}$, $y^{\prime}=-\frac{y}{2}$.
From $M P \perp P Q$, we get
$$
\begin{array}{l}
\frac{y^{\prime}-0}{0-(-3)} \cdot \frac{y^{\prime}}{0-x^{\prime}}=-1 \\
\Rightarrow y^{\prime 2}=3 x^{\prime} \Rightarrow\left(-\frac{y}{2}\right)^{2}=3 \cdot \frac{x}{3} .
\end{array}
$$

Thus, $y^{2}=4 x$ is the trajectory $C$ of point $N$ (excluding the point $(0,0))$.
(2) Let $l: y=k(x+1)(k \neq 0)$.
Substituting into $y^{2}=4 x$, we get
$$
\begin{array}{l}
k^{2} x^{2}+2\left(k^{2}-2\right) x+k^{2}=0 . \\
\text { From } \Delta=\left[2\left(k^{2}-2\right)\right]^{2}-4 k^{2} \cdot k^{2} \\
=-16 k^{2}+16>0,
\end{array}
$$

we get $|k|>1$.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
|A B|=\sqrt{1+k^{2}} \cdot \frac{\sqrt{\Delta}}{k^{2}} \\
=\frac{4}{k^{2}} \sqrt{\left(1+k^{2}\right)\left(1-k^{2}\right)} .
\end{array}
$$

Assume there exists a point $D\left(x_{0}, 0\right)$ such that $\triangle A B D$ is an equilateral triangle.

The midpoint of $A B$ is $E\left(\frac{2-k^{2}}{k^{2}}, \frac{2}{k}\right)$, so the perpendicular bisector of side $A B$ is
$$
y-\frac{2}{k}=-\frac{1}{k}\left(x-\frac{2-k^{2}}{k^{2}}\right) \text {. }
$$

Since point $D$ lies on this perpendicular bisector, we have
$$
0-\frac{2}{k}=-\frac{1}{k}\left(x_{0}-\frac{2-k^{2}}{k^{2}}\right) \Rightarrow x_{0}=\frac{2}{k^{2}}+1 \text {. }
$$

Let $d$ be the distance from point $D$ to the line $l$.
By the condition of the equilateral triangle, we have $\frac{\sqrt{3}}{2}|A B|=d$.
Thus, $\frac{2 \sqrt{3} \sqrt{\left(1+k^{2}\right)\left(1-k^{2}\right)}}{k^{2}}=\frac{2 \sqrt{1+k^{2}}}{|k|}$
$\Rightarrow 3\left(1-k^{2}\right)=k^{2} \Rightarrow k= \pm \frac{\sqrt{3}}{2}, x_{0}=\frac{11}{3}$.
Therefore, there exists a point $D\left(\frac{11}{3}, 0\right)$ such that $\triangle A B D$ is an equilateral triangle.