If 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, then k can have

exactly two values

Given lines will be coplanar, if

$\begin{vmatrix}2-1 & 3-4 & 4-5\\1 & 1 & -k\\k & 2 & 1\end{vmatrix}= 0 \begin{bmatrix} Using\begin{vmatrix}x_2-x_1 & y_2-y_1 & z_2-z_1\\l_1 & m_1 & n_1\\l_2 & m_2 & n_2\end{vmatrix} =0\end{bmatrix}$

$⇒ (1+2k) + (1 + k^2)- (2 - k)= 0 ⇒ k^2 + 3k = 0 ⇒ k = 0 , -3 $