It's pretty easy to get $x=24x'$, $y=30y'$ and $z=40z'$. After that, either $x'$ or $z'$ (but not both) is a multiple of $4$, the other one of these and $y'$ are both odd. $3$ does not divide any of $x'$, $y'$ or $z'$. $5$ must divide either $y'$ or $z'$, but not $x'$. Obviously none of $x'$, $y'$ and $z'$ can have any factors other that $2$, $3$ and $5$, so this forms their complete factorisation. It is easy to verfiy that the two choices are independent, so we get a total of four solutions $\begin{array}{lll} \{x,y,z\} & = & \{24,30,800\} \\ & or & \{24,150,160\}\\& or &\{96,30,200\}\\& or &\{96,150,40\}\end{array}$ corresponding to $\begin{array}{lll} \{x',y',z'\} & = & \{1,1,20\} \\ & or & \{1,5,4\}\\& or &\{4,1,5\}\\& or &\{4,5,1\}.\end{array}$

That's right! However, there is still the condition that $x<y<z$. Therefore there are only 2 solutions.

It's pretty easy to get $x=24x'$, $y=30y'$ and $z=40z'$. After that, either $x'$ or $z'$ (but not both) is a multiple of $4$, the other one of these and $y'$ are both odd. $3$ does not divide any of $x'$, $y'$ or $z'$. $5$ must divide either $y'$ or $z'$, but not $x'$. Obviously none of $x'$, $y'$ and $z'$ can have any factors other that $2$, $3$ and $5$, so this forms their complete factorisation. It is easy to verfiy that the two choices are independent, so we get a total of four solutions $\begin{array}{lll} \{x,y,z\} & = & \{24,30,800\} \\ & or & \{24,150,160\}\\& or &\{96,30,200\}\\& or &\{96,150,40\}\end{array}$ corresponding to $\begin{array}{lll} \{x',y',z'\} & = & \{1,1,20\} \\ & or & \{1,5,4\}\\& or &\{4,1,5\}\\& or &\{4,5,1\}.\end{array}$