Find all solutions of the equation  $$p ^ 2 + q ^ 2 + 49r ^ 2 = 9k ^ 2-101$$ with $ p$, $q$ and $r$ positive prime numbers and $k$ a positive integer.
1. We start with the given Diophantine equation:
   \[
   p^2 + q^2 + 49r^2 = 9k^2 - 101
   \]
   where \( p, q, r \) are positive prime numbers and \( k \) is a positive integer.

2. First, we consider the case where \( p, q, r \) are different from 3. Since \( p, q, r \) are primes, they must be congruent to either 1 or 2 modulo 3. However, the squares of these primes will be congruent to 1 modulo 3:
   \[
   p^2 \equiv q^2 \equiv r^2 \equiv 1 \pmod{3}
   \]
   Therefore,
   \[
   p^2 + q^2 + 49r^2 \equiv 1 + 1 + 49 \cdot 1 \equiv 51 \equiv 0 \pmod{3}
   \]
   This implies that \( 9k^2 - 101 \equiv 0 \pmod{3} \), which simplifies to:
   \[
   9k^2 \equiv 101 \pmod{3} \implies 0 \equiv 2 \pmod{3}
   \]
   This is a contradiction, so at least one of \( p, q, r \) must be 3.

3. Assume \( p = 3 \). Substituting \( p = 3 \) into the original equation, we get:
   \[
   3^2 + q^2 + 49r^2 = 9k^2 - 101 \implies 9 + q^2 + 49r^2 = 9k^2 - 101 \implies q^2 + 49r^2 = 9k^2 - 110
   \]
   This implies:
   \[
   q^2 + 49r^2 \equiv 9k^2 - 110 \pmod{3}
   \]
   Since \( 9k^2 \equiv 0 \pmod{3} \) and \( 110 \equiv 2 \pmod{3} \), we have:
   \[
   q^2 + 49r^2 \equiv -2 \pmod{3} \implies q^2 + r^2 \equiv 1 \pmod{3}
   \]
   This implies that either \( q = 3 \) or \( r = 3 \).

4. If \( q = 3 \), substituting \( q = 3 \) into the equation, we get:
   \[
   3^2 + 49r^2 = 9k^2 - 110 \implies 9 + 49r^2 = 9k^2 - 110 \implies 49r^2 = 9k^2 - 119
   \]
   This implies:
   \[
   9k^2 = 49r^2 + 119
   \]
   Since \( 49r^2 \equiv 1 \pmod{3} \), we have:
   \[
   9k^2 \equiv 1 + 119 \equiv 0 \pmod{3}
   \]
   This is a contradiction, so \( q \neq 3 \).

5. Therefore, \( r = 3 \). Substituting \( r = 3 \) into the equation, we get:
   \[
   p^2 + q^2 + 49 \cdot 3^2 = 9k^2 - 101 \implies p^2 + q^2 + 441 = 9k^2 - 101 \implies p^2 + q^2 = 9k^2 - 542
   \]
   This implies:
   \[
   p^2 + q^2 = 9k^2 - 542
   \]
   We can rewrite this as:
   \[
   (3k - q)(3k + q) = 19 \cdot 29
   \]
   The factor pairs of 551 are \( (1, 551) \) and \( (19, 29) \).

6. If \( (3k - q, 3k + q) = (1, 551) \), then:
   \[
   3k - q = 1 \quad \text{and} \quad 3k + q = 551
   \]
   Adding these equations, we get:
   \[
   6k = 552 \implies k = 92
   \]
   Subtracting these equations, we get:
   \[
   2q = 550 \implies q = 275
   \]
   Since 275 is not a prime, this pair is not valid.

7. If \( (3k - q, 3k + q) = (19, 29) \), then:
   \[
   3k - q = 19 \quad \text{and} \quad 3k + q = 29
   \]
   Adding these equations, we get:
   \[
   6k = 48 \implies k = 8
   \]
   Subtracting these equations, we get:
   \[
   2q = 10 \implies q = 5
   \]
   Since 5 is a prime, this pair is valid.

8. Therefore, the solutions are:
   \[
   (p, q, r, k) = (3, 5, 3, 8) \quad \text{and} \quad (p, q, r, k) = (5, 3, 3, 8)
   \]

The final answer is \( \boxed{ (p, q, r, k) = (3, 5, 3, 8) \text{ and } (5, 3, 3, 8) } \).