Let S be the set of all real values of k for which the system of linear equations x+y+z=2, 2x+y-z=3, 3x+2y+kz=4 has a unique solution. Then S is

The correct option is B Equal to R-0
Given are the system of linear equations.
x+y+z=2
2x+y-z=3
3x+2y+kz=4
We know that, this system has a unique solution.
Therefore, coefficient determinant is non-zero.
|1 1 1|
|2 1 -1| ≠ 0
|3 2 k|
=> k + 2 - (2k + 3) + 1 ≠ 0
=> k ≠ 0
Therefore, k ∈ R - 0 = S