Solving this $4$ variable diophantine equation Is there a way to solve this Diophantine equation in $a,b,c,d$? $$19a^3-33a^2b+3a^2c+30a^2d+21ab^2+24abc-12abd-15ac^2-54acd-30ad^2+ $$ $$2b^3-12b^2c-6b^2d+42bc^2+108bcd+60bd^2-7c^3-51c^2d-99cd^2-56d^3=0$$
Wolfram Alpha unfortunately cannot understand my input whenever I input this equation, and I know no strategy to solve these kinds of equations. So basically, I'm stuck...

I want just all possible values for $a,b,c,d$ just like what Wolfram Alpha does.
Note: $a\neq b\neq c\neq d\neq 0$

Your equation is homogeneous of degree $3$, so if $(a,b,c,d)$ is a solution so is $(ta,tb,tc,td)$ for any $t$.  Thus it makes sense to look for primitive solutions, which are solutions with greatest common divisor $1$.  I found some $288$ primitive solutions with $a,b,c \in [-20\ldots -1, 1\ldots 20]$ and $d \in [1 \ldots 20]$.  I don't see an obvious pattern.
Here are some of those solutions:
$$ \matrix{ a & b & c & d\cr
-4 & -1 & -8 & 1\cr
-4 & 1 & -8 & 3\cr
-3 & 1 & -7 & 3\cr
-3 & 1 & 3 & 1\cr
-3 & 5 & -1 & 4\cr
-2 & 1 & -6 & 3\cr
-2 & 1 & -4 & 2\cr
-2 & 2 & -4 & 3\cr
-2 & 3 & -4 & 4\cr
-2 & 3 & 2 & 1\cr
-2 & 4 & -4 & 5\cr
-2 & 5 & -4 & 6\cr
-1 & -1 & -5 & 4\cr
-1 & 1 & -5 & 3\cr
-1 & 2 & -3 & 3\cr
-1 & 5 & 1 & 1\cr
1 & -2 & -1 & 2\cr
1 & -1 & -5 & 2\cr
1 & 1 & -3 & 3\cr
1 & 2 & -7 & 6\cr
2 & 1 & -2 & 3\cr
2 & 2 & 4 & 1\cr
2 & 3 & 4 & 2\cr
2 & 4 & 4 & 3\cr
2 & 5 & 4 & 4\cr
2 & 6 & 4 & 5\cr
3 & -4 & -7 & 1\cr
3 & 1 & -1 & 3\cr
3 & 2 & -5 & 6\cr
3 & 4 & -3 & 4\cr
4 & 3 & 8 & 1\cr
5 & 1 & 1 & 3\cr
5 & 2 & -3 & 6\cr
6 & 1 & 2 & 3\cr
7 & 1 & 3 & 3\cr
7 & 2 & -1 & 6\cr
8 & 1 & 4 & 3\cr
}$$
Did those coefficients come from somewhere in particular or are they just arbitrary?
