The lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ are coplanar, if

$k = 0 $ or -3

We know that the lines

$\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$

are coplanar, iff

$\begin{vmatrix}x_2-x_1 & y_2-y_1 & z_2-z_1\\l_1 & m_1 & n_1\\l_2 & m_2 & n_2\end{vmatrix}=0$

So, the given lines will be coplanar iff

$\begin{vmatrix}1-2 & 4-3 & 5-4\\1 & 1 & -k\\k & 2 & 1\end{vmatrix}= 0 ⇒ k^2 + 3k = 0 ⇒k = 0, -3$