Since, Only Natural Numbers Solution are needed, So I thought this as a Diophantine Equation, But was unable to handle, 2nd Degrees 3 variable Diophantine Equation.
I Tried to manipulate this Equation as Sum of Squares But I was unsuccessful in doing so, $(2x -7y +3z)^2+16xy-15yz-12xz-4y^2=25$
Can anyone think of some approach ?
Note: Through Online Diophantine Editor, I Found out that it has about 333 integer solutions, But I have a constraint of Natural Numbers and $x<y<z$
On
Slightly more friendly than the suggestion of @Anurag:
Treat the quadratic of $x,y,z$ as a quadratic of $x$ only, then get $$x=\frac{1}{2}[3y \pm \sqrt{25-36y^2-9z^2+36yz}]\implies 2x=3y\pm\sqrt{25-(6y-3z)^2}$$ For the radical to be a perfect square there are three possibilities $(6y-3z)=0,\pm 3, \pm 4,\pm 5.$ You may carry it over now.
I have this until the moment, maybe I'm missing something but I think this can help you
$$ 4x^2 + 45y^2+9z^2-12xy-36yz = 25 $$ $$ 4(x^2-3xy +9y²) +9(y²-4yz+z²) = 25 $$
Let's rename the terms $x² -3xy+9y²$ and $y^2-4yz+z²$ as $s$ and $t$. We know that as x, y and z must be naturals, then $s$ and $t$ must be integers. So you can start by finding all integers that solve the equation $4s + 9t = 25$. A couple comes to mind, $s = 4$ and $t = 1$.
I know this isn't a full solution but maybe it can help you