Explanation:
(B) : It is given that,
\(\mathrm{kx}+2 \mathrm{y}-\mathrm{z}=1,(\mathrm{k}-1) \mathrm{y}-2 \mathrm{z}=2,(\mathrm{k}+2) \mathrm{z}=0\)
For unique solution,
\(\left|\begin{array}{ccc} \mathrm{k} & 2 & -1 \\ 0 & \mathrm{k}-1 & -2 \\ 0 & 0 & \mathrm{k}+2 \end{array}\right| \neq 0\)
\(\mathrm{k}(\mathrm{k}+2)(\mathrm{k}-1) \neq 0\)
\(\mathrm{k}(\mathrm{k}-1)(\mathrm{k}+2) \neq 0\)
\(\mathrm{k} \neq 0,1,-2\)
So, \(\mathrm{k}\) can take any value except \(0,1,-2\).