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 $a= 71, b=9,c=2$
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
3 & 1 & 0 \\
- \frac{ 11 }{ 4 } & - \frac{ 9 }{ 4 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - 3 & - 4 \\
- 3 & 1 & - 6 \\
- 4 & - 6 & 16 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 3 & - \frac{ 11 }{ 4 } \\
0 & 1 & - \frac{ 9 }{ 4 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & - 8 & 0 \\
0 & 0 & \frac{ 81 }{ 2 } \\
\end{array}
\right)
$$
