Let $n$ be the number obtained on rolling a fair die. If the probability that the system
$$
\begin{aligned}
& x-\mathrm{n} y+z=6 \\
& x+(\mathrm{n}-2) y+(\mathrm{n}+1) z=8 \\
& (\mathrm{n}-1) y+z=1
\end{aligned}
$$
has a unique solution is $\frac{k}{6}$, then the sum of $k$ and all possible values of $n$ is :
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