Write $p=a b c, q=\lfloor a\rfloor\lfloor b\rfloor\lfloor c\rfloor$. Note that $q$ is an integer. Multiplying the three equations gives: $$p=\sqrt{\frac{60}{q}}$$ Substitution into the first equation, $$p=3 \frac{a}{\lfloor a\rfloor}<3 \frac{\lfloor a\rfloor+1}{\lfloor a\rfloor} \leq 6$$ Looking at the last equation: $$p=5 \frac{c}{\lfloor c\rfloor} \geq 5 \frac{\lfloor c\rfloor}{\lfloor c\rfloor} \geq 5$$ Here we've used $\lfloor x\rfloor \leq x<\lfloor x\rfloor+1$, and also the apparent fact that $\lfloor a\rfloor \geq 1$. Now: $$\begin{gathered} 5 \leq \sqrt{\frac{60}{q}} \leq 6 \\ \frac{12}{5} \geq q \geq \frac{5}{3} \end{gathered}$$ Since $q$ is an integer, we must have $q=2$. Since $q$ is a product of 3 positive integers, we must have those be 1,1 , and 2 in some order, so there are three cases: Case 1: $\lfloor a\rfloor=2$. By the equations, we'd need $a=\frac{2}{3} \sqrt{30}=\sqrt{120 / 9}>3$, a contradiction, so there are no solutions in this case. Case 2: $\lfloor b\rfloor=2$. We have the solution $$\left(\frac{\sqrt{30}}{3}, \frac{\sqrt{30}}{2}, \frac{\sqrt{30}}{5}\right)$$ Case 3: $\lfloor c\rfloor=2$. We have the solution $$\left(\frac{\sqrt{30}}{3}, \frac{\sqrt{30}}{4}, \frac{2 \sqrt{30}}{5}\right)$$