Answer
The given system has an infinite number of solutions if and only if $k=\frac{1}{3}$ or $k=-4$
Work Step by Step
Determinant for the given matrix is:
$\begin{bmatrix}
1 & 2 &k\\
2 &-k &1\\
0& 0& 1-3k
\end{bmatrix} = \begin{bmatrix}
1 & 2 &k\\
2 &-k &1\\
3& 6& 1
\end{bmatrix}$
$\rightarrow D=(1-3k)(-4-k)=(3k-1)(4+k)$
The given system has an infinite number of solutions if and only if $\det A=0$
Hence, $3k-1=0$ or $4+k=0$
$k=\frac{1}{3}$ or $k=-4$
