there is a simple procedure for finding all rational vectors in the null cone, given $q(z) = 0.$ Let $$ 2 q(x) = x^T H x, $$
where $H$ is the Hessian matrix, all integers. Next, take any primitive integer vector $v$ (we need to assume that $q(v) \neq 0$ and find the value of real $t$ such that $q(z + tv) = 0.$ It turns out that
$$ t = \frac{-2z^T H v}{v^T H v} $$
which is the reason that we needed $q(v) \neq 0.$
Now we have a rational null vector
$$ \frac{(v^THv)z -2(z^T Hv)v}{v^T Hv} $$
To get an integer null vector we may now just multiply by the denominator $q(v).$ The trouble is that the result will often be an integer vector that is not primitive, the GCD of the entries is something larger than one. Well, find the gcd and divide out by it.
So, take column vector $v,$ if $q(v) \neq 0$ find
$$ (v^THv)z -2(z^T Hv)v $$
and divide through by its gcd of entries.
In low dimension there are ways to get around the gcd business. the familiar example would be Pythagorean triples, where primitive null vectors are found parametrized by three binary quadratic forms. I can be specific in dimension 3: given $H$ the Hessian of an integer isotropic form, and the matrix $W$ as the Hessian of $y^2 - zx,$ there is an invertible integer matrix $P$ such that $P^T HP = n W.$ Then recipes can be given for all null vectors.
Fricke and Klein (1897), in dimension 4, use $x^2 + y^2 + z^2 - w^2$ for signature $+++-,$ then $xy - zw$ for signature $++--.$
Clearly, it is enough to decide whether their is rational solutions or not (multiply by the square of a common denominator to have integer solutions).
I assume that $a,b,c,d$ are all nonzero (otherwise, you are reduced to known cases).
Multiplying by $a$ and replacing $x$ by $ax$, one may assume WLOG that $a=1$, so we are reduced to the equation $x^2+by^2+cz^2+dz^2=0$. (*)
This step is not necessary, but it allows to write down simpler conditions.
Hasse Minkowski says that (*) has a nonzero rational solution if and only if it has a nonzero solution over $\mathbb{R}$ and $\mathbb{Q}_p$ for all $p$.
Having a solution over the reals is equivalent to say that $b,c,d$ are not all $>0$.
For the $p$-adic case, it depends on the determinant and local Hasse invariants of the rational quadratic form $x^2+by^2+cz^2+dt^2$.
Here, the determinant is the square class of $bcd$, and if $p$ is prime , the local Hasse invariant is $(b,cd)_p(c,d)_p$.
I will below how to define $( , )_p$.
If $r,s$ are two non zero rationals, write $r=p^\alpha u, s=p^\beta v, p\nmid u, p\nmid v$.
Then $(r,s)_p=(-1)^{\alpha\beta\cdot\frac{p-1}{2}}\left(\dfrac{u}{p}\right)^\beta \left(\dfrac{v}{p}\right)^\alpha$, if $p\neq 2$, where $\left(\dfrac{\phantom{a}}{p}\right)$ is the Legendre symbol,
and $\displaystyle (r,s)_2=(-1)^{\frac{u-1}{2}\cdot \frac{v-1}{2}+\alpha\frac{v^2-1}{8}+\beta\frac{u^2-1}{8}}$
Note for later use that $(-1,-1)_p=1$ if $p\neq 2$ and $(-1,-1)_2=-1$.
Note also that if $p\neq 2$, and the $p$-adic valuations of $r,s$ are both zero, $(r,s)_p=1.$
Recall also the following fact.
Fact. Let $r=p^\alpha u, p\nmid u$. Then $r$ is a square in $\mathbb{Q}_p$ if and only if :
If we translate Thm 6 of Chapter IV $\S$ 2 of Serre's "A course in arithmetic", we get that (*) has a nonzero solution over $\mathbb{Q}_p$ if and only if one of the following cases hold:
Case 1. $bcd$ is not a square modulo in $\mathbb{Q}_p^\times$
Case 2. $bcd$ is a square in $\mathbb{Q}_p^\times$ and $(b,cd)_p=(c,d)_p$ if $p\neq 2$
Case 3. $bcd$ is a square in $\mathbb{Q}_2^\times$ and $(b,cd)_2=-(c,d)_2$
All these conditions amounts to computations of finitely many Legendre symbols, because if $p\nmid bcd$ and $p\neq 2$, then either $bcd$ is not a square in $\mathbb{Q}_p^\times $, or $bcd$ is a square in $\mathbb{Q}_p^\times$ but in this case both symbols $(b,cd)_p$ and $(c,d)_p$ are equal to $1$ !!
Hence for odd prime which are not divisors of $bcd$, you are automatically in one of the first two cases.
So you only have to test if you are in one of the three previous cases only for $p\mid 2bcd $. Note that if the $p$-adic valuation of $bcd$ is odd, $bcd$ is automatically not a square, so you can reduce to prime numbers $p$ such that $p\mid 2bcd$ and $v_p(bcd)$ is even !
Hence, you have an algorithm to decide the existence of a nontrivial solution over $\mathbb{Q}$ (note that the Legendre symbol coincide with the Jacobi symbol, so you can compute it without factoring you integers). However, it won't give you an explicit solution.
Example. Consider $3x^2+3\cdot 5y^2+7z^2-2\cdot 7\cdot 23 t^2=0$.
This is equivalent to consider $x^2+3^2 \cdot 5y^2+3\cdot 7z^2-2\cdot 3 \cdot 7\cdot 23 t^2=0$.
Here $b=3^2 \cdot 5, c=3\cdot 7, d=-2\cdot 3\cdot 7 \cdot 23, cd=-2\cdot 3^2\cdot 7^2\cdot 23$ , and $bcd=-2\cdot 3^4\cdot 5\cdot 7^2\cdot 23$.
Since $b,c,d$ are not all positive, we have solutions over the reals.
For the $p$-adic case, we just have to check two cases: $p=3,7$ since they are the only prime divisors of $bcd$ with an even valuation.
For $p=3$, we need to check that $-2\cdot 5\cdot 23$ is a square modulo $3$.
But $-2\cdot 5\cdot 23=1 \ [3]$, which is a square.
Now since $( r, s)_p$ only depends on the square classes of $r$ and $s$, we have
$(b,cd)_3= (5,-2\cdot 23)_3=(-1)^{\frac{5-1}{2}\frac{-46-1}{2}}=1$ and
$(c,d)_3=(3\cdot 7,-2\cdot 3\cdot 7\cdot 23)_3 =(-1)^{\frac{3-1}{2}}\left(\dfrac{7}{3}\right) \left(\dfrac{-2\cdot 7\cdot 23}{3}\right)=-1\cdot 1\cdot -1=1.$
Hence we are in Case 2.
For $p=7$, we need to check that $-2\cdot 5\cdot 23$ is a square modulo $7$.
But $-2\cdot 5\cdot 23=1 \ [7]$, which is a square.
Since $7\nmid 5,2$ and $3$, $(b,cd)_7=1$. Now $(c,d)_7=(3\cdot 7,-2\cdot 3\cdot 7\cdot 23)_7=(-1)^{\frac{7-1}{2}}\left(\dfrac{3}{7}\right) \left(\dfrac{-2\cdot 3\cdot 23}{7}\right)=-1\cdot -1\cdot 1=1.$
Hence we are in Case 2.
All in all, the original equation must have a non trivial solution.
This is indeed the case since $3\cdot 2^2+3\cdot 5\cdot 3^2+7\cdot 5^2-2\cdot 7\cdot 23\cdot 1^2=0$.
Best Answer
ADDED: it all works. In Dickson's METN, we find, theorem 117 on page 164, that $w^2 + u^2 - 6 z^2$ is the only equivalence class of "determinant" $-6,$ attributed to Ross. Then on page 170, exercise 2, we see that $w^2 + u^2 - 6 z^2$ integrally represents all numbers other than $$ 4^k \left( 16 m + 6 \right) \; , \; \; 9^k \left( 9 m + 6 \right) \; . $$ In particular, $9-6 = 3,$ and all other primes $p \equiv 3 \pmod 8$ are so represented.
ORIGINAL:
This is provisional, I will think about it overnight. The form $$ w^2 - 2 y^2 + 3 y^2 - 6 z^2 $$ is anisotropic in $\mathbb Q_3$ three-adics. It is elementary to show that the form (integrally) represents the product of any two integers it integrally represents, that it does represent $-1,$ and that it integrally represents all (positive) primes $$ p \equiv 1,5,7 \pmod 8 $$
A computer search strongly suggests that it also represents all primes $p \equiv 3 \pmod 8.$ If that can be proved, we are done. Note tht the output below has $x=y,$ meaning that we are looking at an indefinite ternary $w^2 + v^2 - 6 z^2.$