If among numbers $a$, $b$, $c$, $d$ there are two that give either remainders $0$ and $0$ or remainders $1$ and $2$ modulo $3$, the sum of these two numbers is divisible by $3$. Hence $u$, as well as $uv$, is divisible by $3$.

Now study the case where at most one among the numbers $a$, $b$, $c$, $d$ is divisible by $3$ and all numbers not divisible by $3$ are congruent modulo $3$. If exactly one among numbers $a$, $b$, $c$, $d$ is divisible by $3$ then the products of this number with all other numbers are divisible by $3$. Other numbers form $3$ pairs whose products of components are congruent modulo $3$. Hence the sum $v$ of all six pairwise products is divisible by $3$.

If none of $a$, $b$, $c$, $d$ is divisible by $3$ then the pairwise products are all congruent modulo $3$. Again, as the number of pairs is divisible by $3$, this implies that the sum $v$ of the products is divisible by $3$. Consequently, $uv$ is divisible by $3$ in this case, too.