Find Solutions To Some Diophantine Equations I would like to find the solutions to 
$$i)\qquad a(a+b+c)=bc \\ii)\qquad a(a+b+c)=2bc \\iii)\qquad a(a+b+c)=3bc$$
for $0< a \le b \le c$ and of course: $\textrm{gcd}(a,b,c) = 1$ (since those are the interesting cases).
To be honest, Diophantine equations seem like a black art. Guess and check has always been the only sure method that I can understand. 
At least the last case ($iii$) seems to be "easy" as the only solution is apparently $ a = b= c=1$. I don't see how to prove this though.
For $i$, I've found solutions such as: $\{\{1,2,3\}, \{2,3,10\}, \{3,4,21\}, \{3,5,12\}, \ldots\}$
And For $ii$, I've found solutions such as: $\{\{1,1,2\}, \{4,5,6\}, \{15,20,21\}, \{20,22,35\}, \ldots\}$
Notwithstanding my ability to exhaust over integers, I don't have any deeper understanding of what is going on here mathematically. I would really like to understand these better and would appreciate some help! Thanks in advance!
 A: One can give a parameterization to,
$$a(a+b+c) = nbc\tag{1}$$
However, since you want $0<a<b<c$, this makes it trickier.
For $n=1$: A look at your solutions and one can see that $a,b$ are in arithmetic progression. Hence,
$$a,\;b,\;c = n,\;n+1,\;2n^2+n$$
For $n=2$: Most of your $b,c$ differ by one. Hence,
$$a,\;b,\;c = 2pq,\;p^2+pq-1,\;p^2+pq$$
where $p,q$ solve the Pell equation $p^2-3q^2=1$. Since $pq<p^2$, this guarantees that $a<b$.
For $n\ge3$: I checked this with Mathematica and, curiously, there doesn't seem to be any.
P.S. Without the inequalities, then $(1)$ has a two-parameter solution,
$$a,\;b,\;c = (-u + n v)u, \; (u + v)u, \; (-u + n v)v$$
Edit: This answers the comment below. First, we do the substitution $b,\;c = x+y,\;x-y\,$, on $(1)$ then solve for $a$,
$$a =-x\pm\sqrt{(n+1)x^2-ny^2}$$
Thus we have to solve, as W. Jagy pointed out, a ternary quadratic form,
$$(n+1)x^2-ny^2=z^2$$
and initial solution of which is $x,y,z =1,1,1$. Given an initial solution, we generally can find an infinite more. Two methods are described here in Form 2b.
