For $f(-7)$, I see that $-7$ is less than $-5$, so I use the piece $f(x) = 3$. That means $f(-7) = 3$. For $f(0)$, I see that $0$ is between $-5$ and $5$, so I use the piece $f(x) = 2x-3$. That means $f(0) = 2(0) - 3 = -3$. For $f(7)$, I see that $7$ is greater than $5$, so I use the piece $f(x) = x^2+1$. That means $f(7) = 7^2 + 1 = 50$. I get $f(-7)+f(0)+f(7) = 3 + (-3) + 50 = 50$. 
Let \[f(x) = \left\{
\begin{array}{cl}
x^2+1 &\text{ if }x>5, \\
2x-3 &\text{ if } -5 \le x \le 5, \\
3 &\text{ if } x <-5.
\end{array}
\right.\]Find $f(-7)+f(0)+f(7)$.