 Determine the increasing geometric progressions, with three integer terms, such that the sum of these terms is $57$
Given the problem, we need to determine the increasing geometric progressions with three integer terms such that the sum of these terms is \(57\).

1. Let the second term of the geometric progression be \(b\) and the common ratio be \(\frac{p}{q}\) where \((p, q) = 1\). The terms of the geometric progression can be written as:
   \[
   \frac{qb}{p}, \quad b, \quad \frac{bp}{q}
   \]
   Since these terms are integers, \(q \mid b\) and \(p \mid b\). Therefore, \(pq \mid b\), and we can write \(b = apq\) for some integer \(a\). The sequence then becomes:
   \[
   aq^2, \quad apq, \quad ap^2
   \]

2. The sum of these terms is given by:
   \[
   aq^2 + apq + ap^2 = 57
   \]
   Factoring out \(a\), we get:
   \[
   a(p^2 + pq + q^2) = 57
   \]
   We need to find positive integer solutions to this equation with \((p, q) = 1\) and \(p > q\).

3. Since \(a(p^2 + pq + q^2) = 57\), the factor \(p^2 + pq + q^2\) must be a divisor of 57. The divisors of 57 are 1, 3, 19, and 57. We consider the cases where \(a\) and \(p^2 + pq + q^2\) take these values.

4. Case 1: \(a = 1\) and \(p^2 + pq + q^2 = 57\)
   - Clearly, \(q \leq 3\) because if \(q \geq 4\), then \(p^2 + pq + q^2 \geq 4^2 + 4p + p^2 > 57\).

   - Subcase 1.1: \(q = 1\)
     \[
     p^2 + p = 56 \implies p^2 + p - 56 = 0
     \]
     Solving this quadratic equation:
     \[
     p = \frac{-1 \pm \sqrt{1 + 4 \cdot 56}}{2} = \frac{-1 \pm 15}{2}
     \]
     Thus, \(p = 7\) (since \(p\) must be positive). This gives the progression:
     \[
     (1, 7, 49)
     \]

   - Subcase 1.2: \(q = 2\)
     \[
     p^2 + 2p + 4 = 57 \implies p^2 + 2p - 53 = 0
     \]
     Solving this quadratic equation:
     \[
     p = \frac{-2 \pm \sqrt{4 + 4 \cdot 53}}{2} = \frac{-2 \pm 14}{2}
     \]
     Thus, \(p = 6\) (since \(p\) must be positive). This gives the progression:
     \[
     (4, 12, 36)
     \]

   - Subcase 1.3: \(q = 3\)
     \[
     p^2 + 3p + 9 = 57 \implies p^2 + 3p - 48 = 0
     \]
     Solving this quadratic equation:
     \[
     p = \frac{-3 \pm \sqrt{9 + 4 \cdot 48}}{2} = \frac{-3 \pm 15}{2}
     \]
     Thus, \(p = 6\) (since \(p\) must be positive). This gives the progression:
     \[
     (9, 18, 36)
     \]

5. Case 2: \(a = 3\) and \(p^2 + pq + q^2 = 19\)
   - Subcase 2.1: \(q = 1\)
     \[
     p^2 + p = 18 \implies p^2 + p - 18 = 0
     \]
     Solving this quadratic equation:
     \[
     p = \frac{-1 \pm \sqrt{1 + 4 \cdot 18}}{2} = \frac{-1 \pm 9}{2}
     \]
     Thus, \(p = 4\) (since \(p\) must be positive). This gives the progression:
     \[
     (3, 12, 48)
     \]

   - Subcase 2.2: \(q = 2\)
     \[
     p^2 + 2p + 4 = 19 \implies p^2 + 2p - 15 = 0
     \]
     Solving this quadratic equation:
     \[
     p = \frac{-2 \pm \sqrt{4 + 4 \cdot 15}}{2} = \frac{-2 \pm 8}{2}
     \]
     Thus, \(p = 3\) (since \(p\) must be positive). This gives the progression:
     \[
     (12, 18, 27)
     \]

   - Subcase 2.3: \(q = 3\)
     \[
     p^2 + 3p + 9 = 19 \implies p^2 + 3p - 10 = 0
     \]
     Solving this quadratic equation:
     \[
     p = \frac{-3 \pm \sqrt{9 + 4 \cdot 10}}{2} = \frac{-3 \pm 7}{2}
     \]
     Thus, \(p = 2\) (since \(p\) must be positive). This gives the progression:
     \[
     (9, 18, 36)
     \]

Hence, the only solutions are \((1, 7, 49)\) and \((12, 18, 27)\).

The final answer is \( \boxed{ (1, 7, 49) } \) and \((12, 18, 27)\).