To solve the problem, we aim to find the minimum value of the expression

\[
S = a^2 - a + 3b + \frac{9}{a} + \frac{1}{b} + 9
\]

subject to the constraints \(a > 0\), \(b > 0\), and \(2a + b \geq 7\).

### Step 1: Analyze the function components

Let's rewrite \(S\) by grouping the terms related to \(a\) and \(b\):

\[
S = \left(a^2 - a + \frac{9}{a}\right) + \left(3b + \frac{1}{b}\right) + 9
\]

### Step 2: Minimize \(f(a)\) and \(g(b)\)

1. **Minimize \(f(a) = a^2 - a + \frac{9}{a}\)**:

To find the minimum, we differentiate \(f(a)\):
\[
f'(a) = 2a - 1 - \frac{9}{a^2}
\]
Setting \(f'(a) = 0\):
\[
2a - 1 - \frac{9}{a^2} = 0 \quad \Rightarrow \quad 2a^3 - a^2 - 9 = 0
\]

Finding a numerical or analytical solution for this cubic equation might be complicated. Check for a few values:

For \(a = 3\):
\[
f(3) = 3^2 - 3 + \frac{9}{3} = 9 - 3 + 3 = 9
\]

For \(a = 2\):
\[
f(2) = 2^2 - 2 + \frac{9}{2} = 4 - 2 + 4.5 = 6.5
\]

For \(a = 1\):
\[
f(1) = 1^2 - 1 + \frac{9}{1} = 1 - 1 + 9 = 9
\]

Since \(f(a)\) appears to decrease then increase, and from our checks looks minimal around \(a = 2\) yielding \(f(2) = 6.5\).

2. **Minimize \(g(b) = 3b + \frac{1}{b}\)**:

Similarly, minimize \(g(b)\):
\[
g'(b) = 3 - \frac{1}{b^2}
\]
Setting \(g'(b) = 0\):
\[
3 - \frac{1}{b^2} = 0 \quad \Rightarrow \quad b^2 = \frac{1}{3} \quad \Rightarrow \quad b = \frac{1}{\sqrt{3}} \approx 0.577
\]

However, to satisfy \(2a + b \geq 7\) with \(a = 2\):
\[
2(2) + b \geq 7 \quad \Rightarrow \quad 4 + b \geq 7 \quad \Rightarrow \quad b \geq 3
\]

Evaluating \(g(b)\):
For \(b = 3\):
\[
g(3) = 3(3) + \frac{1}{3} = 9 + \frac{1}{3} = 9.33
\]

### Step 3: Combine Results within the Constraints

Check with \(a = 2\) and \(b = 3\):

With these values:
\[
S = 6.5 + 9.33 + 9 = 24.83
\]

### Conclusion

Considering the constraints and tests performed, if we stay with \(a = 2\) and \(b = 3\):
The minimum value of \(S\) is approximately \(24.83\).

Thus, the minimum value of \(S\) is

\[
\boxed{24.83}
\]

under the given constraints.