Identifying relation between numbers based on equation relating them 
If $a^2 + b^2+16c^2=2(3ab+6bc + 4ac)$ , where $a,b,c$ are non zero numbers. Then $a,b,c$ are in __________?

1. Harmonic progression 2. Geometric progression 3. Arithmetic progression 4. None of these
My attempt:
$ a \rightarrow 4a'$
$ b \rightarrow 4b'$
$a'^2 +b'^2 +c^2 - 6a'b' -3b'c -2a'c=0$
Ok, this looks nicer than the original thing but still the solution doesn't seem in sight. Further, I started wondering, would the relationship between numbers be preserved under transformations to the equation?
 A: Try $a=b=4$ and $c=\frac{5+\sqrt{41}}{2}.$
We got: non one of them.
A: There is a way to make this appear simpler. You are using letters a,b,c. Alright, take any $x,y,z$  you like such that
$2x^2 - 16 y^2 + 81 z^2 = 0.$ This is just a point on a cone. Then still to avoid fractions, let
$$ a = 4x+12y - 11z \; , \; \; b = 4y - 9 z \; , \; \; c = 4z $$
For infinitely many examples, we can introduce variables $u,v$ and take
$$ x = 576 u^2 + 72uv \; , \; \; y = 594 u^2 + 144 uv + 9 v^2 \; , \; \; z = 248 u^2 + 64 uv + 4 v^2  $$
For instance, let $u=1, v=1$ to get $x=648,y=747, z=316,$ then $a=8080, b= 144, c = 1264.$  These are all multiples of $16,$ we can divide out to get $a = 505, b= 9, c=79$
Here is a good one, as a single step:
$$ a = 16u^2 + 36 uv + 19 v^2 \; , \; \; b = 9 v^2 \; , \; \; c = 4 u^2 - 2 v^2  $$
With $u=0, v=1$   we get $a=19,b=9, c=-2,$ or with $u=1,v=1$ we get $a= 71, b=9,c=2$
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
3 & 1 & 0 \\ 
 -  \frac{ 11 }{ 4 }  &  -  \frac{ 9 }{ 4 }  & 1 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  - 3 &  - 4 \\ 
 - 3 & 1 &  - 6 \\ 
 - 4 &  - 6 & 16 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 & 3 &  -  \frac{ 11 }{ 4 }  \\ 
0 & 1 &  -  \frac{ 9 }{ 4 }  \\ 
0 & 0 & 1 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
0 &  - 8 & 0 \\ 
0 & 0 &  \frac{ 81 }{ 2 }  \\ 
\end{array}
\right) 
$$
A: Hint:
$$a^2-2a(3b+4c)+b^2+16c^2-12bc=0$$
$$\implies a=\dfrac{2(3b+4c)\pm\sqrt{32b^2+96bc}}2$$
For values of $a,$ we just need $b(b+3c)\ge0$
In that at least one of the two values of $a$ will be $\ne0$
