If the system of linear equations $x+y+z=2, 2x+y-z=3$ and $3x+2y+kz= 4$ has a unique solution, then:

$k≠0$

The correct answer is Option (3) → $k≠0$

Given system of linear equations,

$x+y+z=2$

$2x+y-z=3$

$3x+2y+kz= 4$

$A=\begin{bmatrix}1&1&1\\2&1&-1\\3&2&k\end{bmatrix}$

$det(A)≠0$

$\begin{vmatrix}1&1&2\\2&1&-1\\3&2&k\end{vmatrix}=1(k+2)-1(2k+3)+1(4-3)$

$=k+2-2k-3+1$

$=-k$

$⇒-k≠0$

$⇒k≠0$