If $f(x, y) = 3x^2 + 3xy + 1$ and $f(a, b) + 1 = f(b, a) = 42$, then determine $|a + b|$.
Given the function \( f(x, y) = 3x^2 + 3xy + 1 \) and the conditions \( f(a, b) + 1 = f(b, a) = 42 \), we need to determine \( |a + b| \).

1. Start by writing the given conditions in terms of the function:
   \[
   f(a, b) + 1 = 42 \implies f(a, b) = 41
   \]
   \[
   f(b, a) = 42
   \]

2. Substitute \( f(a, b) \) and \( f(b, a) \) into the function:
   \[
   f(a, b) = 3a^2 + 3ab + 1 = 41 \implies 3a^2 + 3ab + 1 = 41 \implies 3a^2 + 3ab = 40
   \]
   \[
   f(b, a) = 3b^2 + 3ba + 1 = 42 \implies 3b^2 + 3ab + 1 = 42 \implies 3b^2 + 3ab = 41
   \]

3. We now have the system of equations:
   \[
   3a^2 + 3ab = 40
   \]
   \[
   3b^2 + 3ab = 41
   \]

4. Subtract the first equation from the second:
   \[
   3b^2 + 3ab - (3a^2 + 3ab) = 41 - 40
   \]
   \[
   3b^2 - 3a^2 = 1
   \]
   \[
   b^2 - a^2 = \frac{1}{3}
   \]
   \[
   (b - a)(b + a) = \frac{1}{3}
   \]

5. Let \( s = a + b \) and \( d = b - a \). Then:
   \[
   sd = \frac{1}{3}
   \]

6. From the first equation \( 3a^2 + 3ab = 40 \):
   \[
   3a(a + b) = 40 \implies 3a s = 40 \implies a s = \frac{40}{3}
   \]

7. From the second equation \( 3b^2 + 3ab = 41 \):
   \[
   3b(a + b) = 41 \implies 3b s = 41 \implies b s = \frac{41}{3}
   \]

8. Divide the two results:
   \[
   \frac{a s}{b s} = \frac{\frac{40}{3}}{\frac{41}{3}} \implies \frac{a}{b} = \frac{40}{41}
   \]

9. Substitute \( a = \frac{40}{41} b \) into \( s = a + b \):
   \[
   s = \frac{40}{41} b + b = \left( \frac{40}{41} + 1 \right) b = \frac{81}{41} b
   \]

10. Substitute \( a = \frac{40}{41} b \) into \( sd = \frac{1}{3} \):
    \[
    \left( \frac{81}{41} b \right) d = \frac{1}{3} \implies d = \frac{1}{3} \cdot \frac{41}{81 b} = \frac{41}{243 b}
    \]

11. Substitute \( d = b - a \):
    \[
    b - \frac{40}{41} b = \frac{41}{243 b} \implies \frac{1}{41} b = \frac{41}{243 b} \implies b^2 = \frac{41^2}{243} \implies b = \pm \frac{41}{9 \sqrt{3}} = \pm \frac{41 \sqrt{3}}{27}
    \]

12. Calculate \( a \):
    \[
    a = \frac{40}{41} b = \pm \frac{40}{41} \cdot \frac{41 \sqrt{3}}{27} = \pm \frac{40 \sqrt{3}}{27}
    \]

13. Calculate \( |a + b| \):
    \[
    |a + b| = \left| \pm \frac{40 \sqrt{3}}{27} + \pm \frac{41 \sqrt{3}}{27} \right| = \left| \frac{81 \sqrt{3}}{27} \right| = 3 \sqrt{3}
    \]

The final answer is \( \boxed{3 \sqrt{3}} \)