11. Given two lines $I_{1}: 3 x+4 y-25=0, I_{2}: 117 x-44 y-175=0$, point $A$ has projections $B, C$ on lines $I_{1}, l_{2}$, respectively. (1) Find the locus curve $\Gamma$ of point $A$ such that $S_{\triangle A B C}=\frac{1728}{625}$; if circle $T:\left(x-\frac{39}{5}\right)^{2}+\left(y-\frac{27}{5}\right)^{2}=r^{2}(r>0)$ intersects curve $\Gamma$ at exactly 7 points, find the value of $r$.
11. Analysis: (1) Let $A(p, q)$, then $A B=\frac{|3(p-3)+4(q-4)|}{5}, A C=\frac{|117(p-3)-44(q-4)|}{125}$. Let the inclination angles of the lines $\boldsymbol{I}_{1}, \boldsymbol{l}_{2}$ be $\alpha, \beta$, respectively, then the sine of the angle between the lines $\boldsymbol{I}_{1}, \boldsymbol{l}_{2}$ is $\frac{24}{25}$; hence $\frac{1728}{625}=S_{\triangle A B C}=\frac{1}{2} A B \times A C \times \frac{24}{25}=\frac{12}{5^{6}}\left|351(p-3)^{2}-176(q-4)^{2}+336(p-3)(q-4)\right|$, which means the trajectory equation of point $\mathrm{A}$ is: $351(x-3)^{2}-176(y-4)^{2}+336(x-3)(y-4)= \pm 3600$
(2) It is easy to know that the intersection point of the lines $\boldsymbol{I}_{1}, \boldsymbol{l}_{\mathbf{2}}$ is $D(3,4)$. Translate $\mathbf{e} \boldsymbol{T}$ and the curve $\Gamma$ so that point $\mathrm{D}$ is moved to the origin, denoted as point $\mathrm{E}$.
Then, perform a rotation transformation centered at point $\mathrm{E}$, making the angle bisector of $\boldsymbol{I}_{1}, \boldsymbol{l}_{2}$ parallel to the coordinate axes, resulting in $\mathrm{e} \boldsymbol{T}^{*}$ and the curve $\Gamma^{*}$. The coordinate rotation formula is:
$$
\left\{\begin{array} { l } 
{ x ^ { \prime } = \frac { 2 4 } { 2 5 } ( x - 3 ) + \frac { 7 } { 2 5 } ( y - 4 ) } \\
{ y ^ { \prime } = \frac { 2 4 } { 2 5 } ( y - 4 ) - \frac { 7 } { 2 5 } ( x - 3 ) }
\end{array} \Rightarrow \left\{\begin{array}{l}
x-3=\frac{24 x^{\prime}-7 y^{\prime}}{25} \\
y-4=\frac{7 x^{\prime}+24 y^{\prime}}{25}
\end{array}\right.\right.
$$

Substitute the above into the trajectory equation of point $\mathrm{A}$ to get: $16 x^{\prime 2}-9 y^{\prime 2}= \pm 144$, i.e., $\frac{x^{\prime 2}}{9}-\frac{y^{\prime 2}}{16}= \pm 1$, which means the trajectory is two sets of hyperbolas.
Also, $T^{*}\left(\mathbf{5 , 0 )}\right.$, so $\mathrm{e} T^{*}:(x-5)^{2}+y^{2}=r^{2}$; since $\mathrm{e} T$ intersects the curve $\Gamma$ at exactly 7 points, $\mathrm{e} T^{*}$ intersects the curve $\Gamma^{*}$ at exactly 7 points. Since $\mathbf{e} \boldsymbol{T}^{*}$ and the curve $\Gamma^{*}$ are symmetric about the $x$-axis, $(3,0),(-3,0)$ must be on $\mathbf{e} \boldsymbol{T}^{*}$, so $r=2$ or $r=8$.
If $r=2$, then the distance from $\boldsymbol{T}^{*}\left(\mathbf{5 , 0}\right)$ to the asymptote is 4, so $\mathbf{e} \boldsymbol{T}^{*}$ and $\frac{x^{\prime 2}}{9}-\frac{y^{\prime 2}}{16}=-1$ have no intersection points, and at this time $\mathbf{e} \boldsymbol{T}^{*}$ and $\frac{x^{2}}{9}-\frac{y^{\prime 2}}{16}=1$ have at most 4 intersection points, which is a contradiction, so $r=8$.