10. (15 points) Given two lines
$$
\begin{array}{l}
l_{1}: 3 x+4 y-25=0, \\
l_{2}: 117 x-44 y-175=0,
\end{array}
$$

Point $A$ has projections $B$ and $C$ on $l_{1}$ and $l_{2}$, respectively.
(1) Find the locus curve $\Gamma$ of point $A$ such that $S_{\triangle A B C}=\frac{1728}{625}$;
(2) If $\odot T:\left(x-\frac{39}{5}\right)^{2}+\left(y-\frac{27}{5}\right)^{2}=r^{2}$ intersects the curve $\Gamma$ at exactly 7 points, find the value of $r$.
10. (1) Let $A(x, y)$. It is easy to know
$$
\begin{array}{l}
A B=\frac{|3 x+4 y-25|}{\sqrt{3^{2}+4^{2}}}=\frac{|3(x-3)+4(y-4)|}{5}, \\
A C=\frac{|117 x-44 y-175|}{\sqrt{117^{2}+44^{2}}} \\
=\frac{|117(x-3)-44(y-4)|}{125} .
\end{array}
$$

Let the angles of inclination of $l_{1}$ and $l_{2}$ be $\dot{\alpha}_{1}$ and $\alpha_{2}$, respectively, and the angle between $l_{1}$ and $l_{2}$ be $\beta=\alpha_{1}-\alpha_{2}$. Then $\sin \beta=\frac{24}{25}$. Therefore,
$$
\begin{aligned}
& S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \sin \beta \\
= & \frac{12}{5}\left|351(x-3)^{2}-176(y-4)^{2}+336(x-3)(y-4)\right| \\
= & \frac{1728}{625} \\
\Rightarrow & \mid 351(x-3)^{2}-176(y-4)^{2}+ \\
& 336(x-3)(y-4) \mid=3600 .
\end{aligned}
$$

Equation (1) is the trajectory equation of point $A$.
(2) It is easy to know $T\left(\frac{39}{5}, \frac{27}{5}\right)$, and the intersection point of $l_{1}$ and $l_{2}$ is $D(3,4)$.

Translate $\odot T$ and curve $\Gamma$ so that point $D$ is moved to the origin, and then perform a rotation transformation centered at point $D^{\prime}$ to make the angle bisector of $l_{1}$ and $l_{2}$ parallel to the coordinate axes, obtaining $\odot T^{\prime}$ and curve $\Gamma^{\prime}$, i.e.,
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{\prime}=\frac{24}{25}(x-3)+\frac{7}{25}(y-4), \\
y^{\prime}=\frac{24}{25}(y-4)-\frac{7}{25}(x-3)
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x-3=\frac{24 x^{\prime}-7 y^{\prime}}{25}, \\
y-4=\frac{7 x^{\prime}+24 y^{\prime}}{25} .
\end{array}\right.
\end{array}
$$

Substitute into equation (1) and simplify to get
$$
\left|\left(4 x^{\prime}\right)^{2}-\left(3 y^{\prime}\right)^{2}\right|=144 \text {. }
$$

Then curve $\Gamma^{\prime}$ is two sets of hyperbolas:
$$
\Gamma_{1}: \frac{x^{2}}{9}-\frac{y^{2}}{16}=1, \Gamma_{2}: \frac{x^{2}}{9}-\frac{y^{2}}{16}=-1 \text {. }
$$

Also, $T^{\prime}(5,0)$, so $\odot T^{\prime}:(x-5)^{2}+y^{2}=r^{2}$.
Since $\odot T$ intersects curve $\Gamma$ at exactly 7 points, i.e., $\odot T^{\prime}$ intersects $\Gamma^{\prime}$ at exactly 7 points, and both $\odot T$ and $\Gamma^{\prime}$ are symmetric about the $x$-axis, $\odot T$ and $\Gamma^{\prime}$ must have one intersection point on the $x$-axis, i.e., $(-3,0)$ or $(3,0)$ is on $\odot T^{\prime}$. Therefore,
$r=2$ or 8.
If $r=2$, since the distance from $T^{\prime}(5,0)$ to the asymptotes of $\Gamma^{\prime}$ is
$$
\frac{|4 \times 5 \pm 3 \times 0|}{\sqrt{4^{2}+3^{2}}}=4>r,
$$

so, $\odot T^{\prime}$ does not intersect $\Gamma_{2}$. And $\odot T^{\prime}$ intersects $\Gamma_{1}$ at most 4 points, which contradicts the fact that $\odot T^{\prime}$ intersects $\Gamma^{\prime}$ at 7 points. Therefore, $r=8$.
It is easy to see that when $r=8$, the conditions are satisfied.