As $\text{gcd}(a, b) = 2$, $\text{gcd}(b, c) = 3$ and $\text{gcd}(c, a) = 5$, the number $a$ is divisible by both $2$ and $5$, the number $b$ is divisible by both $2$ and $3$, and the number $c$ is divisible by both $3$ and $5$. Hence $a$ is divisible by $10$, $b$ is divisible by $6$ and $c$ is divisible by $15$. As $61$ gives remainder $1$ when divided by each of $2$, $3$ and $5$, the numbers $a, b$ and $c$ must give remainder $1$ when divided by $3$, $5$ and $2$, respectively. Since $a, b, c \le 61$, the possibilities are $a = 10$ or $a = 40$, $b = 6$ or $b = 36$, and $c = 15$ or $c = 45$. The sum $61$ appears in three cases: $a = 10, b = 6, c = 45$; $a = 10, b = 36, c = 15$; $a = 40, b = 6, c = 15$. A straightforward check shows that the conditions $\text{gcd}(a, b) = 2$, $\text{gcd}(b, c) = 3$ and $\text{gcd}(c, a) = 5$ are also met in all these cases.