Olympiads Number Theory Problem: Show that there are no positive $a,b,c\in\Bbb Z$ such that $\frac{a^2+b^2+c^2}{3(ab+bc+ca)}\in\Bbb Z$

$\textbf{Hint:}$Assume such positive integers exist.Then,

There exists $n$ such that $a^2+b^2+c^2=3n(ab+bc+ca)$.

Now,add $2(ab+bc+ca)$ on both sides.That would give $(a+b+c)^2=(3n+2)(ab+bc+ca)$

Now,$3n+2$ has a prime divisor say $p$ appearing odd number of times of the form $3k+2$ for some $k$.From here deduce that

$p \mid a+b+c$ and $p \mid ab+bc+ca$.

Deduce from here that, $p \mid a^2+ab+b^2$

After that use quadratic reciprocity to derive some contradiction.

Given integers $B > A > 0,$ with $\gcd(A,B) = 1,$ then there is a solution in integers $x,y,z,$ not all $0,$ to $$ A(x^2 + y^2 + z^2) - B (yz + zx + xy) = 0, $$ if and only if both $B+2A$ and $B-A$ are integrally represented by the binary form $u^2 + 3 v^2.$

In your case $A = 1$ and $B = 3k$ for some integer $k$

My $B+2A = 3k+2$ and cannot be represented by $u^2 + 3 v^2,$ so that is it since your vriables are restricted to positive.

Summary at Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$