Find all $3$-tuples $(a, b, c)$ of positive integers, with $a \geq b \geq c$, such that $a^2 + 3b$, $b^2 + 3c$, and $c^2 + 3a$ are all squares.
1. Given the conditions \(a \geq b \geq c\) and the equations \(a^2 + 3b\), \(b^2 + 3c\), and \(c^2 + 3a\) are all squares, we start by analyzing the first equation \(a^2 + 3b\).

2. Since \(a \geq b\), we have:
   \[
   a^2 + 3b \leq a^2 + 3a < a^2 + 4a + 4 = (a + 2)^2
   \]
   However, \(a^2 < a^2 + 3b\), so:
   \[
   a^2 + 3b = (a + 1)^2 \implies 3b = 2a + 1
   \]

3. Similarly, for the second equation \(b^2 + 3c\):
   \[
   b^2 + 3c = (b + 1)^2 \implies 3c = 2b + 1
   \]

4. Solving these equations, we get:
   \[
   a = \frac{3b - 1}{2} \quad \text{and} \quad b = \frac{3c - 1}{2}
   \]

5. Substituting \(b\) into the equation for \(a\):
   \[
   a = \frac{3 \left(\frac{3c - 1}{2}\right) - 1}{2} = \frac{9c - 5}{4}
   \]
   Since \(a\) is an integer, \(c \equiv 1 \pmod{4}\).

6. Now, we need \(c^2 + 3a\) to be a perfect square. Substituting \(a = \frac{9c - 5}{4}\):
   \[
   c^2 + \frac{27c - 15}{4} \text{ is a perfect square}
   \]
   Multiplying both sides by 4:
   \[
   4c^2 + 27c - 15 \text{ is a perfect square}
   \]

7. We analyze the possible values for \(4c^2 + 27c - 15\):
   \[
   (2c + 8)^2 = 4c^2 + 32c + 64 > 4c^2 + 27c - 15
   \]
   \[
   (2c)^2 < 4c^2 + 27c - 15
   \]
   Since \(4c^2 + 27c - 15\) is even, we have:
   \[
   4c^2 + 27c - 15 \in \{(2c + 2)^2, (2c + 4)^2, (2c + 6)^2\}
   \]

8. Case 1: \(4c^2 + 27c - 15 = (2c + 2)^2\):
   \[
   4c^2 + 27c - 15 = 4c^2 + 8c + 4 \implies 19c = 19 \implies c = 1
   \]
   So, we get the solution:
   \[
   (a, b, c) = (1, 1, 1)
   \]
   which works.

9. Case 2: \(4c^2 + 27c - 15 = (2c + 4)^2\):
   \[
   4c^2 + 27c - 15 = 4c^2 + 16c + 16 \implies 11c = 31
   \]
   So, \(c\) is not an integer, contradiction.

10. Case 3: \(4c^2 + 27c - 15 = (2c + 6)^2\):
    \[
    4c^2 + 27c - 15 = 4c^2 + 24c + 36 \implies 3c = 51 \implies c = 17
    \]
    So, we get the solution:
    \[
    (a, b, c) = (37, 25, 17)
    \]
    which works.

The final answer is \( \boxed{ (1, 1, 1) } \) and \((37, 25, 17)\).