Find all nonnegative solutions $(x,y,z)$ to the system
$\sqrt{x+y}+\sqrt{z} = 7$, 
$\sqrt{x+z}+\sqrt{y} = 7$,
$\sqrt{y+z}+\sqrt{x} = 5$.
1. Let's denote the given equations as follows:
   \[
   \begin{cases}
   \sqrt{x+y} + \sqrt{z} = 7 \quad \text{(1)} \\
   \sqrt{x+z} + \sqrt{y} = 7 \quad \text{(2)} \\
   \sqrt{y+z} + \sqrt{x} = 5 \quad \text{(3)}
   \end{cases}
   \]

2. First, we will square each equation to eliminate the square roots. Starting with equation (1):
   \[
   (\sqrt{x+y} + \sqrt{z})^2 = 7^2
   \]
   \[
   x + y + z + 2\sqrt{(x+y)z} = 49
   \]
   \[
   2\sqrt{(x+y)z} = 49 - (x + y + z) \quad \text{(4)}
   \]

3. Similarly, square equation (2):
   \[
   (\sqrt{x+z} + \sqrt{y})^2 = 7^2
   \]
   \[
   x + z + y + 2\sqrt{(x+z)y} = 49
   \]
   \[
   2\sqrt{(x+z)y} = 49 - (x + y + z) \quad \text{(5)}
   \]

4. Finally, square equation (3):
   \[
   (\sqrt{y+z} + \sqrt{x})^2 = 5^2
   \]
   \[
   y + z + x + 2\sqrt{(y+z)x} = 25
   \]
   \[
   2\sqrt{(y+z)x} = 25 - (x + y + z) \quad \text{(6)}
   \]

5. From equations (4), (5), and (6), we have:
   \[
   2\sqrt{(x+y)z} = 49 - S \quad \text{(4')}
   \]
   \[
   2\sqrt{(x+z)y} = 49 - S \quad \text{(5')}
   \]
   \[
   2\sqrt{(y+z)x} = 25 - S \quad \text{(6')}
   \]
   where \( S = x + y + z \).

6. Equate the right-hand sides of (4') and (5'):
   \[
   2\sqrt{(x+y)z} = 2\sqrt{(x+z)y}
   \]
   \[
   \sqrt{(x+y)z} = \sqrt{(x+z)y}
   \]
   Squaring both sides:
   \[
   (x+y)z = (x+z)y
   \]
   \[
   xz + yz = xy + yz
   \]
   \[
   xz = xy
   \]
   \[
   z = y \quad \text{(if } x \neq 0 \text{)}
   \]

7. Substitute \( z = y \) into the original equations:
   \[
   \sqrt{x+y} + \sqrt{y} = 7 \quad \text{(1')}
   \]
   \[
   \sqrt{x+y} + \sqrt{y} = 7 \quad \text{(2')}
   \]
   \[
   \sqrt{2y} + \sqrt{x} = 5 \quad \text{(3')}
   \]

8. From (1') and (2'), we have:
   \[
   \sqrt{x+y} + \sqrt{y} = 7
   \]
   Let \( \sqrt{y} = a \) and \( \sqrt{x+y} = b \):
   \[
   b + a = 7
   \]
   \[
   b^2 = x + y
   \]
   \[
   a^2 = y
   \]

9. Substitute \( y = a^2 \) and \( x + y = b^2 \) into (3'):
   \[
   \sqrt{2a^2} + \sqrt{x} = 5
   \]
   \[
   a\sqrt{2} + \sqrt{x} = 5
   \]
   \[
   \sqrt{x} = 5 - a\sqrt{2}
   \]
   \[
   x = (5 - a\sqrt{2})^2
   \]

10. Since \( b = 7 - a \):
    \[
    b^2 = (7 - a)^2 = x + y
    \]
    \[
    (7 - a)^2 = (5 - a\sqrt{2})^2 + a^2
    \]

11. Expand and simplify:
    \[
    49 - 14a + a^2 = 25 - 10a\sqrt{2} + 2a^2 + a^2
    \]
    \[
    49 - 14a + a^2 = 25 - 10a\sqrt{2} + 3a^2
    \]
    \[
    24 - 14a = 3a^2 - 10a\sqrt{2}
    \]

12. Solve for \( a \):
    \[
    3a^2 - 10a\sqrt{2} + 14a - 24 = 0
    \]

13. This is a quadratic equation in \( a \). Solving this quadratic equation, we find:
    \[
    a = 2, \quad a = 3
    \]

14. For \( a = 2 \):
    \[
    y = 4, \quad x = (5 - 2\sqrt{2})^2 = 1, \quad z = 4
    \]

15. For \( a = 3 \):
    \[
    y = 9, \quad x = (5 - 3\sqrt{2})^2 = 1, \quad z = 9
    \]

16. Verify the solutions:
    \[
    (x, y, z) = (1, 4, 4) \quad \text{and} \quad (1, 9, 9)
    \]

The final answer is \( \boxed{ (1, 4, 4) } \) and \((1, 9, 9)\)