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

Given lines are

$\frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{k}$

and  $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$

Since the lines are coplanar, so 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\left|\begin{array}{ccc}1& -1& -1\\ 1& 1& -k\\ k& 2& 1\end{array}\right|=0\\ 1(1+2\mathrm{k})+1\left(1+{\mathrm{k}}^2\right)-1(2-\mathrm{k})=0\\ \Rightarrow 1+2\mathrm{k}+{\mathrm{k}}^2+1-2+\mathrm{k}=0\\ \Rightarrow {\mathrm{k}}^2+3\mathrm{k}=0\\ \Rightarrow \mathrm{k}(\mathrm{k}+3)=0\end{array}$

$\Rightarrow \mathrm{k}=0$ or $\mathrm{k}=-3$