Answer: \(k=-4\). For this value of \(k\), the solutions are \(x=t\), \(y=3t-2\), \(z=\frac74t-\frac54\), where \(t\in\mathbb R\).

For the system not to have a unique solution, the determinant of the coefficient matrix must be zero:

\(\begin{vmatrix}2&-3&4\\3&-1&0\\1&2&k\end{vmatrix}=0\).

Expanding along the first row gives

\(2\begin{vmatrix}-1&0\\2&k\end{vmatrix}+3\begin{vmatrix}3&0\\1&k\end{vmatrix}+4\begin{vmatrix}3&-1\\1&2\end{vmatrix}=0\).

Hence \(-2k+9k+28=0\), so \(7k=-28\) and therefore \(k=-4\).

With \(k=-4\), the first and third equations add to

\((2x-3y+4z)+(x+2y-4z)=1+1\),

so \(3x-y=2\), which is the second equation. Thus one parameter remains free.

Let \(x=t\). Then \(3t-y=2\), so \(y=3t-2\).

Substitute into \(x+2y-4z=1\):

\(t+2(3t-2)-4z=1\).

Thus \(7t-4z=5\), giving \(z=\frac74t-\frac54\).

So the solution set is \(x=t\), \(y=3t-2\), \(z=\frac74t-\frac54\), where \(t\in\mathbb R\).