Let the second number be $b$ and the ratio be $\frac{p}{q}$ with $(p,q)=1$. Then the numbers are $\frac{qb}{p}, b,\frac{bp}{q}$. So $q \mid b$ and $p \mid b$ and so $pq \mid b$ so $b=apq$ for some $a$ and the sequence is given by $aq^2, apq, ap^2$. So we are looking for positive integer solutions to $a(p^2+pq+q^2)=57$ with $(p,q)=1$ and $p>q$. So the second factor is at least $7$ and divides $57=3 \cdot 19$. Hence either $a=1$ and $p^2+pq+q^2=57$ or $a=3$ and $p^2+pq+q^2=19$. We distinguish cases: 1. $p^2+pq+q^2=19$. Case 1.1. $q=1$. Then $p^2+p=18$ and $(2p+1)^2=73$ which is not a square. So no solution in this case. Case 1.2. $q \ge 2$. Then $p \ge 3$ and $p^2+pq+q^2 \ge 3^2+2 \cdot 3+2^2=19$. Hence equality must hold and $q=2, p=3$ which is indeed a solution yielding the progression $(12,18,27)$. 2. $p^2+pq+q^2=57$. Clearly $q \le 3$ since otherwise $p^2+pq+q^2 \ge 5^2+4 \cdot 5+4^2=61>57$. Case 2.1. $q=1$. Then $p^2+p=56$ and so $(2p+1)^2=225$ so $2p+1=15$ and $p=7$ which is indeed a solution yielding the progression $(1,7,49)$. Case 2.2. $q=2$. Then $p^2+2p=53$ and $(p+1)^2=54$ which is not a square. So no solution in this case. Case 2.3. $q=3$. Then $p^2+3p=48$ and $(2p+3)^2=201$ which is not a square. So no solution in this case. Hence the only solutions are $(1,7,49)$ and $(12,18,27)$.

Let $\frac{p}{q}, mdc(p,q) = 1, 1\leq q<p$ (since it is increasing) be the geometrical progression ratio. It needs to be rational, because, otherwise, at least one of the $3$ terms of the progression wouldn't be rational, and so wouldn't be an integer. Let the three numbers be $a,\frac{ap}{q},\frac{p^2a}{q^2}$. Once that $mdc({p},{q})=1 \Rightarrow mdc({p^2},{q^2})=1 \Rightarrow q^2|a$, there exists some integer $b$ such that $a = q^2b$ Substituting $a$ for $q^2b$, the terms of the progression become $p^2b, bpq, q^2b.$ Adding all of them, we have $b\left(p^2 + pq + q^2\right)=3\cdot19.$ Since $1\leq q<p$ $, 7\leq(p^2 + pq + q^2)$. Since $7\leq(p^2 + pq + q^2),(p^2 + pq + q^2)\neq1$ and $(p^2 + pq + q^2)\neq3, $ we have two cases$(p^2 + pq + q^2) = 19$ and $(p^2 + pq + q^2) = 57$ Case 1.0 $(p^2 + pq + q^2)=19$ Case 1.1$p=4,q=1$ give us $21>19$ Case 1.1 $p=3,q=2, (p^2 + pq + q^2)=19$, what we needed. Case 1.2 $3\leq p$ and $ q<2 \Rightarrow (p^2 + pq + q^2)<19$ Case 2.0 $(p^2 + pq + q^2)=57 $ Case 2.1.1 $p \leq 5$ and $q <4 \Rightarrow (p^2 + pq + q^2)<49$ Case 2.1.2 $p=5$ and $q=4 \Rightarrow (p^2 + pq + q^2)>61>57$ Case 2.2.1 $p = 6$ and $q \leq 2 \Rightarrow (p^2 + pq + q^2) \leq 52<57$ Case 2.2.2 $p=6$ and $3 \leq q \Rightarrow (p^2 + pq + q^2) \geq 63 > 57$ Case 2.3.0 $p=7$ and $q = 1 \Rightarrow (p^2 + pq + q^2) = 57$, as we desired Case 2.3.1 $p =7$ and $q>1 \Rightarrow (p^2 + pq + q^2) \geq 64>57$ Case 2.4 $p>7 and q \geq 1 \Rightarrow (p^2 + pq + q^2) \geq 73>57$ So the only cases are $(p,q) = (3,2), (7,1).$ And so the possible ratios are $\frac{3}{2}$ and $\frac{7}{1}$. Case 1: the ratio is $\frac{3}{2}: a + \frac{3a}{2} + \frac{9a}{4} = 57 \Leftrightarrow a =12, \frac{3a}{2} = 18, \frac{9a}{4} =27$ Case 2: the ratio is $\frac{7}{1}: a + 7a + 49a = 57 \Leftrightarrow a = 1, 7a=7 ,49a = 49 $ So the answer is $(a,ar,ar^2) = (12,18,27), (1,7,49)$

The question didn't said that the three terms are consecutive so I think that the sequence (3,6,48) also works. The three sequences are all the solutions.