We have the following congruences:
$a^2+2dbc \equiv 0 \mod 3$
$b^2+2ac \equiv 0 \mod 3$
$2ab+dc^2 \equiv 0 \mod 3$
where $a,b,c \in \mathbb{Z}$ and $d$ is a square free integer that is not divisible by $3$ and $d \not \equiv \pm 1 \mod 9$.
Deduce that $a \equiv c \equiv \pm 1 \mod 3$ and $b \equiv dc \mod 3$.
I don't get why these should be solutions. Moreover $a \equiv b \equiv c \equiv 0 \mod 3$ is a solution too.