Find all the natural solutions of (a+b+c)a-3bc=0 I was trying to solve a geometry puzzle when I came across a simple algebraic problem that I couldn't solve.

Given the expression $(a+b+c)a - 3bc = 0$, find all natural solutions for $a$, $b$ and $c$.

I've tried to isolate one variable, like $$a=\frac{-b-c\pm\sqrt{b^2+14bc +c^2}}{2}$$  but I didn't get anywhere. Despite this, I've notice that the numbers $(2, 6, 1)$ satisfy the condition. Does anyone can help with this problem model?
 A: Rewrite the equation as $(a+b)(a+c)=4bc$. Now 
\begin{eqnarray*}
a+b= 2 \alpha \beta \\
a+c = 2 \gamma \delta \\
b= \alpha \gamma \\
c= \beta \delta
\end{eqnarray*}
will satisfy this provided $ \alpha ( 2 \beta - \gamma)=a= \delta(2 \gamma - \beta)$. So choose 
\begin{eqnarray*}
\alpha=2 \gamma - \beta \\
\delta= 2 \beta - \gamma
\end{eqnarray*}
and so we have faimily of solutions generated by
\begin{eqnarray*}
a= (2 \gamma - \beta)( 2 \beta - \gamma) \\
b= \gamma (2 \gamma - \beta)\\ 
c= \beta ( 2 \beta - \gamma).\\
\end{eqnarray*}
A: $$c=\frac{a^2+ab}{3b-a}$$
The problem then becomes: for which $a,b$ do we have $3b-a|a(a+b)$. 
Let $d= gcd(a,b)$ then $a=da', b=db'$ with $gcd(a',b')=1$. Then
$$3b'-a'|da'(a'+b')$$
Now, let us observe that 
$$gcd(3b'-a', a')|3gcd(a',b')=3$$ and 
$$gcd(3b'-a', a'+b')|4gcd(a',b')=4 \,.$$
At this point the problem reduces to 6 cases, and a case by case analysis will complete the problem:
$$gcd(3b'-a', a') \in \{1,3\} \, \mbox{ and  } \, gcd(3b'-a', a'+b') \in \{ 1,2,4\}$$
But one can also write the general solution this way:
Pick $a',b'$ arbitrary. Let 
$$d=\frac{3b'-a'}{gcd(3b'-a', a')gcd(3b'-a', a'+b')} \cdot l$$
for some integer $l$.
Then 
$$a=da'=\frac{3b'-a'}{gcd(3b'-a', a')gcd(3b'-a', a'+b')} \cdot la'  , \\
b=\frac{3b'-a'}{gcd(3b'-a', a')gcd(3b'-a', a'+b')} \cdot l b', \\ c=\frac{a^2+ab}{3b-a}= \frac{a'(a'+b')}{gcd(3b'-a', a')gcd(3b'-a', a'+b')} \cdot l$$
is a solution, and this describes the general solution.
