what value of K does the system have a unique solution
If $k\neq -\frac12$
$$\left(
\begin{matrix}
1 & k & -1 & | & 2\\
2 & -1 & k & | & 5 \\
1 & 10 & -6 & | & 1 \\
\end{matrix}
\right) \xrightarrow[\text{$R_3=R_3-R_1$}]{\text{$R_2=R_2-2R_1$}}$$ $$\left(
\begin{matrix}
1 & k & -1 & | & 2\\
0 & -1-2k & k+2 & | & 1 \\
0 & 10-k & -5 & | & -1 \\
\end{matrix}
\right) \xrightarrow[\text{$(2k+1\neq0)$}]{R_3=R_3-(\frac{10-k}{2k+1})R_2} $$
$$\left(
\begin{matrix}
1 & k & -1 & | & 2\\
0 & -1-2k & k+2 & | & 1 \\
0 & 0 & \frac{-k^2-2k+15}{2k+1} & | & \frac{-3k+9}{2k+1} \\
\end{matrix}
\right) $$
Which has a unique solution $\iff \frac{-3k+9}{2k+1} \neq 0 \iff -3k+9\neq 0 \iff k\neq3$
If $k=-\frac12$ it is easy to show that the system has a unique solution.