Given a square $ABCD$. Let $P\in{AB},\ Q\in{BC},\ R\in{CD}\ S\in{DA}$ and $PR\Vert BC,\ SQ\Vert AB$ and let $Z=PR\cap SQ$. If $BP=7,\ BQ=6,\ DZ=5$, then find the side length of the square.
1. **Identify the variables and given conditions:**
   - Let the side length of the square be \( s \).
   - Given points: \( P \in AB \), \( Q \in BC \), \( R \in CD \), \( S \in DA \).
   - Given distances: \( BP = 7 \), \( BQ = 6 \), \( DZ = 5 \).
   - Given parallel lines: \( PR \parallel BC \) and \( SQ \parallel AB \).
   - Intersection point: \( Z = PR \cap SQ \).

2. **Express the coordinates of the points:**
   - Since \( P \in AB \) and \( BP = 7 \), the coordinates of \( P \) are \( (7, s) \).
   - Since \( Q \in BC \) and \( BQ = 6 \), the coordinates of \( Q \) are \( (s, s-6) \).
   - Since \( PR \parallel BC \), \( R \) must be directly below \( P \) on \( CD \). Therefore, \( R \) has coordinates \( (7, 0) \).
   - Since \( SQ \parallel AB \), \( S \) must be directly to the left of \( Q \) on \( DA \). Therefore, \( S \) has coordinates \( (0, s-6) \).

3. **Determine the coordinates of \( Z \):**
   - Since \( PR \parallel BC \) and \( SQ \parallel AB \), \( Z \) is the intersection of these lines.
   - The line \( PR \) is vertical, passing through \( (7, s) \) and \( (7, 0) \), so its equation is \( x = 7 \).
   - The line \( SQ \) is horizontal, passing through \( (0, s-6) \) and \( (s, s-6) \), so its equation is \( y = s-6 \).
   - Therefore, the coordinates of \( Z \) are \( (7, s-6) \).

4. **Use the given distance \( DZ = 5 \):**
   - The coordinates of \( D \) are \( (s, 0) \).
   - The distance \( DZ \) is given by:
     \[
     DZ = \sqrt{(7 - s)^2 + (s - 6)^2} = 5
     \]
   - Squaring both sides:
     \[
     (7 - s)^2 + (s - 6)^2 = 25
     \]

5. **Simplify the equation:**
   \[
   (7 - s)^2 + (s - 6)^2 = 25
   \]
   \[
   (s - 7)^2 + (s - 6)^2 = 25
   \]
   \[
   (s - 7)^2 = (s^2 - 14s + 49)
   \]
   \[
   (s - 6)^2 = (s^2 - 12s + 36)
   \]
   \[
   s^2 - 14s + 49 + s^2 - 12s + 36 = 25
   \]
   \[
   2s^2 - 26s + 85 = 25
   \]
   \[
   2s^2 - 26s + 60 = 0
   \]
   \[
   s^2 - 13s + 30 = 0
   \]

6. **Solve the quadratic equation:**
   \[
   s^2 - 13s + 30 = 0
   \]
   - Using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
     \[
     s = \frac{13 \pm \sqrt{169 - 120}}{2}
     \]
     \[
     s = \frac{13 \pm \sqrt{49}}{2}
     \]
     \[
     s = \frac{13 \pm 7}{2}
     \]
     \[
     s = \frac{20}{2} = 10 \quad \text{or} \quad s = \frac{6}{2} = 3
     \]
   - Since \( s > 7 \), we discard \( s = 3 \).

7. **Conclusion:**
   \[
   s = 10
   \]

The final answer is \( \boxed{10} \)