What is the largest prime $p$ that makes $\sqrt{17p+625}$  an integer?

$ \textbf{(A)}\ 3
\qquad\textbf{(B)}\ 67
\qquad\textbf{(C)}\ 101
\qquad\textbf{(D)}\ 151
\qquad\textbf{(E)}\ 211
$
To determine the largest prime \( p \) such that \(\sqrt{17p + 625}\) is an integer, we start by letting \(\sqrt{17p + 625} = q\), where \( q \) is an integer. 

1. **Express the equation in terms of \( q \):**
   \[
   \sqrt{17p + 625} = q \implies 17p + 625 = q^2
   \]
   Rearrange to solve for \( p \):
   \[
   17p = q^2 - 625
   \]
   \[
   p = \frac{q^2 - 625}{17}
   \]

2. **Factorize the quadratic expression:**
   \[
   q^2 - 625 = (q - 25)(q + 25)
   \]
   Thus, we have:
   \[
   17p = (q - 25)(q + 25)
   \]

3. **Analyze the factors:**
   Since \( p \) must be a prime number, and \( 17p \) must be the product of two factors, one of these factors must be 17 (since 17 is a prime number). Therefore, we consider two cases:
   
   - Case 1: \( q - 25 = 1 \)
     \[
     q - 25 = 1 \implies q = 26
     \]
     Substitute \( q = 26 \) into the equation for \( p \):
     \[
     17p = (26 - 25)(26 + 25) = 1 \cdot 51 = 51
     \]
     \[
     p = \frac{51}{17} = 3
     \]
     Here, \( p = 3 \), which is a prime number.

   - Case 2: \( q - 25 = 17 \)
     \[
     q - 25 = 17 \implies q = 42
     \]
     Substitute \( q = 42 \) into the equation for \( p \):
     \[
     17p = (42 - 25)(42 + 25) = 17 \cdot 67
     \]
     \[
     p = 67
     \]
     Here, \( p = 67 \), which is also a prime number.

4. **Determine the largest prime:**
   Comparing the two values of \( p \) obtained, \( p = 3 \) and \( p = 67 \), the largest prime is \( p = 67 \).

Therefore, the largest prime \( p \) that makes \(\sqrt{17p + 625}\) an integer is \( \boxed{67} \).