#### Answer

This system won't have a unique solution for $$k\in\{-2,1\}$$

#### Work Step by Step

To analyze this system easier we will transform it to row echelon form doing the steps described below:
Step 1: Subtract the first equation from the second to eliminate $x$:
$$x-x+ky-y+z-kz = 2-3$$
which becomes
$$(k-1)y-(k-1)z=-1$$
Our system now reads
\begin{align*}
x+y+kz=&3\\
(k-1)y-(k-1)z=-1\\
kx+y+z=1
\end{align*}
Step 2: Subtract the first equation multiplied by $k$ from the third one to loose $x$:
$$kx-kx+y-ky+z-k^2z=1-3k$$
and this becomes
$$-(k-1)y -(k-1)(k+1)z=1-3k$$
where we used the fact that $1-k^2 = (1-k)(1+k) = -(k-1)(k+1)$.
Multiplying by $-1$ we get
$$(k-1)y+(k-1)(k+1)z=3k-1$$
Now our system reads
\begin{align*}
x+y+kz=&3\\
(k-1)y-(k-1)z=&-1\\
(k-1)y+(k-1)(k+1)z=&3k-1
\end{align*}
Step 3: Now subtract the second equation from the third to eliminate $y$:
$$(k-1)y-(k-1)y+(k-1)(k+1)z-(-(k-1)z)=3k-1-(-1)$$
and this becomes
$$(k+2)(k-1)z=3k$$
Our system reads:
\begin{align*}
x+y+kz=&3\\
(k-1)y-(k-1)z=&-1\\
(k+2)(k-1)z=3k.
\end{align*}
Step 4.
Note that if we take $k=1$ the second equation will say
$$(1-1)z-(1-1)y=-1\Rightarrow 0=-1$$
which is impossible so then we will have no solutions.
Also if we take $k=-2$ the last equation will read:
$$(-2+2)(-2-1)z=3\times(-2)\Rightarrow 0=-6$$
which also impossible and we will have no solutions. For any other value of $k$ except these two we can work our way backwards and solve first for $z$ then for $y$ and then for $x$ and find a unique solution. In the problem we are required NOT to have a unique solution and that means we should choose $k\in\{-2,1\}$