For what values k is the system consistent?

I assume you mean $T \colon \mathbb{R}^3 \rightarrow \mathbb{R}^2$.

We have $Im(T) = \mathsf{span}\left( \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \begin{bmatrix} 2 \\ -4 \end{bmatrix}, \begin{bmatrix} 3 \\ -k \end{bmatrix} \right) = \mathsf{span}\left( \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \begin{bmatrix} 3 \\ -k \end{bmatrix} \right)$ as the second vector is a linear multiple of the first.

1. The vector $\begin{bmatrix} -17 \\ 34 \end{bmatrix}$ belongs to the image for every value of $k$, as it is $-17 \cdot \begin{bmatrix} 1 \\ -2 \end{bmatrix}$.

2. Since the vector $\begin{bmatrix} 1 \\ 34 \end{bmatrix}$ is not a linear multiple of $\begin{bmatrix} 1 \\ -2 \end{bmatrix}$, if belongs to the image if and only if the map is surjective, that is, whenever $\begin{bmatrix} 3 \\ -k \end{bmatrix}$ is not a linear multiple of $\begin{bmatrix} 1 \\ -2 \end{bmatrix}$, so whenever $k \neq 6$. In this case, in fact, the matrix has rank $2$ and so $T$ is surjective.

3. As explained in the second point, $T$ is surjective $\iff k \neq 6$.

4. $T$ is a map from a space of higher dimension to a space of lower dimension, so it is never injective, regardless of the value of $k$. Indeed, \[ T \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} = 0 \] for every value of $k$.