To solve the system of equations

\[
\begin{cases}
3(x + 2) + \sqrt{y - 6} = 11 \quad (1) \\
5(x + 2) - \sqrt{y - 6} = 13 \quad (2)
\end{cases}
\]

let's start with these equations.

### Step 1: Solve Equation (1)

From equation (1):
\[
3(x + 2) + \sqrt{y - 6} = 11
\]
Subtract \(3(x + 2)\) from both sides:
\[
\sqrt{y - 6} = 11 - 3(x + 2)
\]
Simplifying gives:
\[
\sqrt{y - 6} = 11 - 3x - 6
\]
\[
\sqrt{y - 6} = 5 - 3x \quad (3)
\]

### Step 2: Solve Equation (2)

From equation (2):
\[
5(x + 2) - \sqrt{y - 6} = 13
\]
Add \(\sqrt{y - 6}\) to both sides:
\[
5(x + 2) - 13 = \sqrt{y - 6}
\]
Simplifying gives:
\[
\sqrt{y - 6} = 5(x + 2) - 13
\]
\[
\sqrt{y - 6} = 5x + 10 - 13
\]
\[
\sqrt{y - 6} = 5x - 3 \quad (4)
\]

### Step 3: Set Equations (3) and (4) Equal

Now we can set equations (3) and (4) equal to each other:
\[
5 - 3x = 5x - 3
\]
Rearranging gives:
\[
5 + 3 = 5x + 3x
\]
\[
8 = 8x
\]
\[
x = 1
\]

### Step 4: Substitute \(x = 1\) Back to Find \(y\)

Substituting \(x = 1\) into one of the earlier equations, we can use equation (3):
\[
\sqrt{y - 6} = 5 - 3(1)
\]
\[
\sqrt{y - 6} = 5 - 3
\]
\[
\sqrt{y - 6} = 2
\]
Now square both sides:
\[
y - 6 = 4
\]
\[
y = 10
\]

### Final Solution

Thus, the solution is:
\[
\boxed{(1, 10)}
\]