Let n be the number obtained on rolling a fair die. If the probability that the system

$$ \begin{aligned} & x-\mathrm{n} y+z=6 \\ & x+(\mathrm{n}-2) y+(\mathrm{n}+1) z=8 \\ & \quad(\mathrm{n}-1) y+z=1 \end{aligned} $$

has a unique solution is $\frac{k}{6}$, then the sum of $k$ and all possible values of $n$ is :

If $X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ is a solution of the system of equations $A X=B$, where $\operatorname{adj} A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$, then $|x+y+z|$ is equal to :

If $\mathrm{A}=\left[\begin{array}{ll}2 & 3 \\ 3 & 5\end{array}\right]$, then the determinant of the matrix $\left(\mathrm{A}^{2025}-3 \mathrm{~A}^{2024}+\mathrm{A}^{2023}\right)$ is

If the system of equations

$ 3x + y + 4z = 3 $

$ 2x + \alpha y - z = -3 $

$ x + 2y + z = 4 $

has no solution, then the value of $ \alpha $ is equal to: