If the trivial solution is the only solution of the system of equations

$x-ky + z=0$

$kx + 3y-kz = 0$

$3x+y-z=0$, then the set of values of k, is

$R - \{2,-3\}$

The given system of equations will have trivial solution only, if

$\begin{vmatrix}1&-k&1\\k&3&-k\\3&1&-1\end{vmatrix}≠0$

$⇒-3+k+k(-k + 3k) + (k −9) ≠0$

$⇒2k^2 + 2k-12≠0$

$⇒k^2+k-6≠0$

$⇒k≠-3,2$

Hence, $k∈R - \{2,-3\}$.