17. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, passing through the point $P\left(3, \frac{16}{5}\right)$, with an eccentricity of $\frac{3}{5}$. A line $l$ with slope $k$ is drawn through the right focus of the ellipse $C$, intersecting the ellipse at points $A$ and $B$. The slopes of $P A$ and $P B$ are denoted as $k_{1}$ and $k_{2}$, respectively.
(1) Find the standard equation of the ellipse;
(2) If $k_{1}+k_{2}=0$, find the real number $k$.
(1) $\frac{9}{a^{2}}+\frac{256}{25 b^{2}}=1 \Rightarrow \frac{9 b^{2}}{a^{2}}=b^{2}-\frac{256}{25} \Rightarrow b^{2}-\frac{256}{25}=9\left[1-\left(\frac{3}{5}\right)^{2}\right]=\frac{144}{25} \Rightarrow b^{2}=16, a^{2}=25$.

So the standard equation of the ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$.
(2) It is easy to verify that when $k=0$, $A(-5,0), B(5,0) \Rightarrow k_{1}+k_{2} \neq 0$, which does not meet the requirement;

Let the line $l: x=n y+3$, where $n=\frac{1}{k}$. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$,
Combining $\left\{\begin{array}{l}x=n y+3, \\ 16 x^{2}+25 y^{2}=400\end{array} \Rightarrow\left(16 n^{2}+25\right) y^{2}+96 n y-256=0\right.$.
Then $k_{1}+k_{2}=0 \Rightarrow \frac{y_{1}-\frac{16}{5}}{x_{1}-3}+\frac{y_{2}-\frac{16}{5}}{x_{2}-3}=0 \Rightarrow\left(y_{1}-\frac{16}{5}\right) y_{2}+\left(y_{2}-\frac{16}{5}\right) y_{1}=0$
$\Rightarrow 2 y_{1} y_{2}-\frac{16}{5}\left(y_{1}+y_{2}\right)=2 \cdot(-256)-\frac{16}{5} \cdot(-96 n)=0 \Rightarrow n=\frac{5}{3} \Rightarrow k=\frac{3}{5}$.