<s>[INST] A Diophantine Equation 
Finding the number of $(a, b, c)$, where $a, b, c$ are positive
  integers, that
$$
\frac{a^2+b^2-c^2}{ab}+\frac{c^2+b^2-a^2}{cb}+\frac{a^2+c^2-b^2}{ac}=2+\frac{15}{abc} 
$$


I factored it in the following form
$(a+b+c)(2ab+2ac+2bc-a^2-b^2-c^2)=8abc+15$. But I can't make any progress from here.
Also I tried working with case when $a, b, c$ are sides of a triangle to get,
$\cos A+\cos C+\cos B=1+\frac{15}{2abc} < \frac{3}{2}$.
But that gives a lower bound on $abc$ which is not very helpful either.
 [/INST] Hint
$$\Longleftrightarrow c(a^2+b^2-c^2)+a(b^2+c^2-a^2)+b(a^2+c^2-b^2)-2abc=15$$
and note that
$$c(a^2+b^2-c^2)+a(b^2+c^2-a^2)+b(a^2+c^2-b^2)-2abc=(c-a+b)(a-b+a)(c-a-b)$$
so
$$(c-a+b)(c-b+a)(c-a-b)=15=1\cdot 3\cdot 5$$
</s>