Determine the natural number $a =\frac{p+q}{r}+\frac{q+r}{p}+\frac{r+p}{q}$ where $p, q$ and $r$ are prime positive numbers.
1. Given the expression \( a = \frac{p+q}{r} + \frac{q+r}{p} + \frac{r+p}{q} \), where \( p, q, r \) are prime numbers, we need to determine the natural number \( a \).

2. Let's rewrite the expression in a common denominator:
   \[
   a = \frac{(p+q)q + (q+r)r + (r+p)p}{pqr}
   \]
   Simplifying the numerator:
   \[
   (p+q)q + (q+r)r + (r+p)p = pq + q^2 + qr + r^2 + rp + p^2
   \]
   Therefore,
   \[
   a = \frac{pq + q^2 + qr + r^2 + rp + p^2}{pqr}
   \]

3. We need to check if \( p, q, r \) are distinct primes. Assume \( p, q, r \) are distinct primes. We need to check if \( pqr \mid pq + q^2 + qr + r^2 + rp + p^2 \).

4. Consider the divisibility conditions:
   - \( p \mid q^2 + qr + r^2 \)
   - \( q \mid r^2 + rp + p^2 \)
   - \( r \mid p^2 + pq + q^2 \)

5. Since \( p, q, r \) are distinct primes, the greatest common divisor (gcd) of any two of them is 1:
   - \( \gcd(p, q) = 1 \)
   - \( \gcd(q, r) = 1 \)
   - \( \gcd(r, p) = 1 \)

6. Given the gcd conditions, we can infer that:
   - \( p \mid q+r \)
   - \( q \mid r+p \)
   - \( r \mid p+q \)

7. Since \( p, q, r \) are distinct primes, the only possible values for \( p, q, r \) that satisfy these conditions are \( p = 2, q = 3, r = 5 \).

8. Substituting \( p = 2, q = 3, r = 5 \) into the expression:
   \[
   a = \frac{2+3}{5} + \frac{3+5}{2} + \frac{5+2}{3}
   \]
   Simplifying each term:
   \[
   \frac{2+3}{5} = 1, \quad \frac{3+5}{2} = 4, \quad \frac{5+2}{3} = 2
   \]
   Therefore,
   \[
   a = 1 + 4 + 2 = 7
   \]

The final answer is \(\boxed{7}\).