#### Answer

For ANY $k\neq 0$ we will have exactly one solution and it will be given by
$$x= -\frac{23}{9},\quad y=-\frac{16}{9};\quad z=\frac{13}{3}.$$

#### Work Step by Step

Firstly we will transform this system to Row-Echelon form in order to analyze it easier following the steps below:
Step 1: Change the order of the equations as follows:
\begin{align*}
x+y+z=&0\\
2x-y+z=&0\\
kx+2ky+3kz=&4k
\end{align*}
Step 2: Subtract from the second equation the first one multiplied by $2$ to eliminate $x$:
$$2x-2x-y-2y+z-2z=1-2\times0$$
which becomes
$$-3y-z=1$$
and multiplied by $-1$ for nicer looks:
$$3y+z=-1$$
Our system is now:
\begin{align*}
x+y+z=&0\\
3y+z=&-1\\
kx+2ky+3kz=&4k
\end{align*}
Step 3: Subtract from the third equation the first one multiplied by $k$ to loose $x$:
$$
kx-kx+2ky-ky+3kz-kz=4k-k\times0
$$
which becomes
$$ky+2kz =4k.$$
Now our system reads
\begin{align*}
x+y+z=&0\\
3y+z=&-1\\
ky+2kz=&4k
\end{align*}
Step 4: Subtract from the third equation the second one multiplied by $\frac{k}{3}$ to loose $y$:
$$ky-\frac{k}{3}3y+2kz-\frac{k}{3}z = 4k-\frac{k}{3}(-1)$$
which becomes
$$\frac{5}{3}kz=\frac{13}{3}k$$
Multiply by $3$ for nicer looks
$$5kz=13k.$$
Now our system reads:
\begin{align*}
x+y+z=&0\\
3y+z=&-1\\
3kz=&13k
\end{align*}
Step 5: Notice that for $k=0$ the third equation won't be an actual equation but will just say $0=0$ and in this cas we would have infinitely many solutions depending on one parameter. For any $k\neq$ we will have exactly one solution because then it is "legal" (because we know $k$ is not zero) to divide the last equation with $k$ and that gives
$$3z=13\Rightarrow z=\frac{13}{3}.$$
Now we can put this into the second equation to get
$$3y+\frac{13}{3} = -1\Rightarrow 3y = -1-\frac{13}{3}\Rightarrow 3y= -\frac{16}{3}$$
and that gives the solution for $y$:
$$y=-\frac{16}{9}.$$
Finally we can put the calculated values for $y$ and $z$ into the first equation and find $x$:
$$x-\frac{16}{9}+\frac{13}{3} = 0\Rightarrow x +\frac{23}{9} = 0$$
and this gives
$$x=-\frac{23}{9}$$
So for ANY $k\neq 0$ we will have exactly one solution and it will be given by
$$x= -\frac{23}{9},\quad y=-\frac{16}{9};\quad z=\frac{13}{3}.$$