Find prime numbers $p , q , r$  such that $p+q^2+r^3=200$. Give all the possibilities.

Remember that the number $1$ is not prime.
1. We start by noting that \( p, q, r \) are prime numbers and we need to find all possible combinations such that \( p + q^2 + r^3 = 200 \). Since 1 is not a prime number, we exclude it from our considerations.

2. Since the sum \( p + q^2 + r^3 \) is even (200 is even), at least one of the primes must be 2 (the only even prime number). We will consider three cases: \( p = 2 \), \( q = 2 \), and \( r = 2 \).

3. **Case 1: \( r = 2 \)**
   \[
   p + q^2 + 2^3 = 200 \implies p + q^2 + 8 = 200 \implies p + q^2 = 192
   \]
   Since \( p \) and \( q \) are primes and cannot both be even, they must both be odd. Also, \( q^2 < 192 \implies q \leq 13 \).

   - If \( q = 3 \):
     \[
     p + 3^2 = 192 \implies p + 9 = 192 \implies p = 183 \quad (\text{not prime})
     \]
   - If \( q = 5 \):
     \[
     p + 5^2 = 192 \implies p + 25 = 192 \implies p = 167 \quad (\text{prime})
     \]
   - If \( q = 7 \):
     \[
     p + 7^2 = 192 \implies p + 49 = 192 \implies p = 143 \quad (\text{not prime})
     \]
   - If \( q = 11 \):
     \[
     p + 11^2 = 192 \implies p + 121 = 192 \implies p = 71 \quad (\text{prime})
     \]
   - If \( q = 13 \):
     \[
     p + 13^2 = 192 \implies p + 169 = 192 \implies p = 23 \quad (\text{prime})
     \]

   So, the solutions for \( r = 2 \) are:
   \[
   (p, q, r) = (167, 5, 2), (71, 11, 2), (23, 13, 2)
   \]

4. **Case 2: \( q = 2 \)**
   \[
   p + 2^2 + r^3 = 200 \implies p + 4 + r^3 = 200 \implies p + r^3 = 196
   \]
   Since \( p \) and \( r \) are primes and cannot both be even, they must both be odd. Also, \( r^3 < 196 \implies r \leq 5 \).

   - If \( r = 3 \):
     \[
     p + 3^3 = 196 \implies p + 27 = 196 \implies p = 169 \quad (\text{not prime})
     \]
   - If \( r = 5 \):
     \[
     p + 5^3 = 196 \implies p + 125 = 196 \implies p = 71 \quad (\text{prime})
     \]

   So, the solution for \( q = 2 \) is:
   \[
   (p, q, r) = (71, 2, 5)
   \]

5. **Case 3: \( p = 2 \)**
   \[
   2 + q^2 + r^3 = 200 \implies q^2 + r^3 = 198
   \]
   Since \( q \) and \( r \) are primes and cannot both be even, they must both be odd. Also, \( r^3 < 198 \implies r \leq 5 \).

   - If \( r = 3 \):
     \[
     q^2 + 3^3 = 198 \implies q^2 + 27 = 198 \implies q^2 = 171 \quad (\text{not an integer})
     \]
   - If \( r = 5 \):
     \[
     q^2 + 5^3 = 198 \implies q^2 + 125 = 198 \implies q^2 = 73 \quad (\text{not an integer})
     \]

   So, there are no solutions for \( p = 2 \).

6. Combining all the valid solutions from the cases above, we have:
   \[
   (167, 5, 2), (71, 11, 2), (23, 13, 2), (71, 2, 5)
   \]

The final answer is \( \boxed{ (167, 5, 2), (71, 11, 2), (23, 13, 2), (71, 2, 5) } \).