Preface: the most natural way to take one isotropic vector for an indefinite quadratic form and find others is to use stereographic projection. This gives a parametrization in the same $n$ variables as the quadratic form. After clearing denominators, there is also not much control over the gcd of the resulting integers. So, although finding an integer multiple of every primitive solution is guaranteed, we may not be entirely sure we have found all primitive solutions with entries up to some bound in absolute value.
There is a trick for indefinite ternary forms, which leads to a parametrization by two parameters, with considerable control of the gcd's.
Most of this appears in my answers to Isotropic ternary forms
Question: is it true that the primitive integer solutions to $$ A(x^2 + y^2 + z^2) - B (yz+zx+xy) =0 $$ can all be parametrized by a finite number of solutions as below, in shorthand $R_j U?$ The calculations are awfully convincing, but I have proved only a few. In case anyone gets interested, i wrote out the proof for $A=2, B=113,$ about twenty four pages pdf.
Once we have integers $B > A > 0,$ a necessary and sufficient condition that the form be isotropic in $\mathbb Q,$ and therefore $\mathbb Z,$ is that both $B-A$ and $B+2A$ have integer expressions as $s^2 + 3 t^2.$
There is an interesting alternative, method goes back to Fricke and Klein, gives a two variable parametrization, and can be adjusted to deal with GCD's. There is a complete answer to this, finding all primitive solutions, meaning $\gcd(x,y,z) = 1.$
We begin with finding all primitive solutions to $y^2 - z x = 0.$ If $g = \gcd(x,z) > 1,$ then $g^2 | y^2$ and $g | y,$ so $g | \gcd(x,y,z).$ However, $\gcd(x,y,z) = 1.$ So, $\gcd(x,z) = 1.$ Since $xz = y^2,$ either $x=u^2, z=v^2,$ or $x=-u^2,z=-v^2,$ in either case with $\gcd(u,v) = 1.$ That is, possibly by changing from $(x,y,z)$ to $(-x,-y,-z)$ so as to arrange $x \geq 0,$ all primitive solutions are $$ x = u^2, y = u v, z = v^2. $$
Next, the quadratic form is $X^T G X / 2,$ where $$ G = \left( \begin{array}{rrr} 2a & d & e \\ d & 2b & f \\ e & f &2c \end{array} \right) $$ and $$ X = \left( \begin{array}{r} x \\ y \\ z \end{array} \right) $$
The quadratic form $y^2 - z x$ is $X^T H X / 2,$ where $$ H = \left( \begin{array}{rrr} 0 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 0 \end{array} \right) $$ It is a theorem in Fricke and Klein (1897), pages 507-508, that, because the quadratic form has at least one integer solution, in which $(x,y,z)$ are not all zero, there exists a square matrix of integers $R$ and a nonzero integer $n$ such that $$ R^T G R = n H. $$ As $R$ has an inverse, we can take $S$ to have integral entries and minimize positive $k$ in $$ RS = SR = k I. $$
We already know that we can take all solutions of $y^2 - z x = 0$ as the column vector $$ U = \left( \begin{array}{r} u^2 \\ uv \\ v^2 \end{array} \right) $$ for relatively prime $(u,v).$ That is, $U^T H U = 0,$ and all solutions are a scalar multiple of $U.$
What happens if $X^T G X = 0,$ the "solutions" we want, with gcd one? Well, $R^T G R = n H,$ so $S^T R^T G R S = n S^T H S,$ so $$ G = \frac{n}{k^2} S^T H S, $$ and $X^T G X = 0$ says $$ X^T S^T H S X = 0. $$ We have already shown that there is some integer $w$ with $$ SX = w U. $$ This gives us $RSX = w RU$ and $kX = w RU.$ Now, as $\gcd(x,y,z) = 1,$ there is a row vector $A = (\alpha,\beta,\gamma)$ with $AX = 1.$ This tells us $k = w ARU.$ As $ARU$ is some integer, $w | k,$ and the earlier $kX = w RU$ becomes $$ X = \frac{1}{h} R U $$ for $h = \frac{k}{w} \in \mathbb Z.$ Furthermore, as $ RS = SR = k I, $ we know $k | \det R,$ so $h | k$ tells us $h | \det R.$ One may leave it this way: list the divisors of $\det R,$ including $\det R$ itself. For each primitive pair $(u,v),$ produce the column vector $RU,$ which will be a solution but perhaps not primitive. Divide out by the gcd of the entries of $RU.$ All integer primitive solutions are given by $$ X = RU/ g_1, $$ where $g_1$ is the gcd of the three entries of $RU.$ It is worth emphasizing that $g_1$ is a divisor of $\det R.$ Also, we get some explicit bounds, as $$ |X|^2 = \frac{1}{g_1^2} U^T R^T R U, $$ since $R$ is nonsingular integer and $R^T R$ is symmetric positive definite. So, no matter what, we have a way to find all primitive solutions $X$ with some $|X| \leq \mbox{bound}$ by taking $|u|, |v|$ up to some other bound we can figure out. $$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$ A more interesting alternative: for each divisor of $\det R,$ we may rewrite the eventual primitive solution with that gcd as a new recipe, $R_1 U$ for a new integer matrix $R_1$ that also solves $R_1^T G R_1 = n H.$
The example I like to show is solving $$ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0, $$ four "recipes," $$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 37 u^2 + 51 uv + 8 v^2 \\ 8 u^2 -35 uv -6 v^2 \\ -6 u^2 + 23 uv + 37 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 32 u^2 + 61 uv + 18 v^2 \\ 18 u^2 -25 uv -11 v^2 \\ -11 u^2 + 3 uv + 32 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 38 u^2 + 45 uv + 4 v^2 \\ 4 u^2 -37 uv -3 v^2 \\ -3 u^2 + 31 uv + 38 v^2 \end{array} \right) $$
$$ \left( \begin{array}{r} x \\ y \\ z \end{array} \right) = \left( \begin{array}{r} 29 u^2 + 63 uv + 22 v^2 \\ 22 u^2 -19 uv -12 v^2 \\ -12 u^2 -5 uv + 29 v^2 \end{array} \right) $$
For all four recipes, $$ x^2 + y^2 + z^2 = 1469 \left( u^2 + uv + v^2 \right)^2 $$ Since $u^2 + uv + v^2 \geq 3 u^2 / 4$ and $u^2 + uv + v^2 \geq 3 v^2 / 4,$ this gives us explicit bounds on the absolute values of $u,v$ that gives us all (primitive) solutions of $ 2(x^2 + y^2 + z^2) = 113 (yz+zx+xy) $ with the absolute values of $x,y,z$ up to a desired bound.
Indeed, we were able to choose all four coefficient matrices with this pattern: $$ R = \left( \begin{array}{ccc} \alpha & \beta & \gamma \\ \gamma & - \beta + 2 \gamma & \alpha - \beta + \gamma \\ \alpha - \beta + \gamma & 2 \alpha - \beta & \alpha \end{array} \right) $$
The rows constitute a cycle of three neighboring, but not reduced, binary quadratic forms under the action of the matrix $$ P = \left( \begin{array}{rr} 0 & 1 \\ -1 & -1 \end{array} \right), $$ where $P^3 = I.$ As soon as we write $X = RU$ we get the identity $$ x^2 + y^2 + z^2 = \left( \alpha^2 + (\alpha - \beta + \gamma)^2 + \gamma^2 \right) \cdot \left( u^2 + uv + v^2 \right)^2 $$
In all four cases we simply discard occurrences when the resulting $x,y,z$ have a common factor. With the understanding that we negate all $x,y,z$ so that the entry with largest absolute value is positive, then sort so that $$ x \geq |y| \geq |z|, $$ here are the answers with maximum up to $1200$
jagy@phobeusjunior:~$ ./isotropy_binaries_combined 2 113 1200 | sort -n
x y z first line u v
29 22 -12 < 29, 63, 22 > 1 0
32 18 -11 < 32, 61, 18 > 1 0
37 8 -6 < 37, 51, 8 > 1 0
38 4 -3 < 38, 45, 4 > 1 0
188 171 -86 < 37, 51, 8 > 1 2
211 144 -82 < 38, 45, 4 > 1 2
226 123 -76 < 32, 61, 18 > 1 2
243 94 -64 < 29, 63, 22 > 1 2
246 88 -61 < 38, 45, 4 > 2 1
258 59 -44 < 37, 51, 8 > 2 1
264 38 -29 < 29, 63, 22 > 2 1
268 11 -6 < 32, 61, 18 > 2 1
396 262 -151 < 37, 51, 8 > 1 3
432 209 -134 < 38, 45, 4 > 1 3
472 129 -94 < 29, 63, 22 > 3 1
489 76 -58 < 32, 61, 18 > 3 1
516 458 -233 < 38, 45, 4 > 2 3
526 447 -232 < 37, 51, 8 > 2 3
628 311 -198 < 38, 45, 4 > 3 2
656 262 -177 < 32, 61, 18 > 2 3
671 232 -162 < 37, 51, 8 > 3 2
692 183 -134 < 29, 63, 22 > 2 3
726 47 -32 < 32, 61, 18 > 3 2
727 36 -22 < 29, 63, 22 > 3 2
804 787 -382 < 32, 61, 18 > 1 5
894 688 -373 < 29, 63, 22 > 1 5
953 946 -456 < 38, 45, 4 > 3 4
1034 492 -317 < 37, 51, 8 > 1 5
1062 443 -296 < 29, 63, 22 > 5 1
1102 363 -256 < 38, 45, 4 > 1 5
1123 314 -228 < 32, 61, 18 > 5 1
1159 1046 -528 < 32, 61, 18 > 1 6
1179 118 -88 < 38, 45, 4 > 5 1
1188 19 2 < 37, 51, 8 > 5 1
1199 1002 -524 < 29, 63, 22 > 1 6
x y z first line u v
I should probably point out that, while it was quite easy (after writing the C++ programs) to make a list of primitive solutions to $ 2(x^2 + y^2 + z^2) - 113(yz + zx + xy)=0 $ with $x \geq |y| \geq |z|$ for $x \leq 1200,$ and just as easy to identify the four square matrices $R_1,R_2,R_3,R_4$ used above, it was quite a big job to prove that these really do give all (ordered) primitive solutions. I have a pdf of the whole business in detail, about twenty pages Latex. Oh: in the above, we may always take $u,v \geq 0.$ It is a reasonable conjecture that the problem $ A(x^2 + y^2 + z^2) - B(yz + zx + xy)=0, $ with $\gcd(A,B)=0,$ $B > A > 0,$ and both $B-A$ and $B + 2A$ expressible in integers as $s^2 + 3 t^2,$ always works out with a finite number of such $R_i.$ No proof.
