Six, since $a$ is a root of the equation, then
$$a^{4}+3 a^{3}+2 a^{2}=2 \times 3^{n} \text {, }$$

which means $a^{2}(a+2)(a+1)=2 \times 3^{n}$
From this, we get $a \neq 0,-1,-2$.
If $a$ is even, then the left side of the equation is a multiple of 4, while the right side is not, leading to a contradiction. Therefore, $a$ is odd.

When $a$ is odd, $a$ and $a+2$ have only one that is a multiple of 3.
If $a$ is a multiple of 3, then
$|a+2|=1 \Rightarrow a=-1$ (discard), $a=-3$.
When $a=-3$, $n=2$.
If $a+2$ is a multiple of 3, then
$|a|=1 \Rightarrow a=-1$ (discard), $a=1$.
When $a=1$, $n=1$.
Therefore, the pairs that satisfy the condition are
$$(n, a)=(2,-3),(1,1) .$$