If the lines $2\mathrm{x}+\mathrm{y}-\mathrm{k}=0$,  $\mathrm{x}-\mathrm{y}+1=0$ and  $-2\mathrm{x}-3\mathrm{y}+8=0$ are concurrent then  k=

The condition that the lines ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1=0$ , ${\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2=0$ and ${\mathrm{a}}_3\mathrm{x}+{\mathrm{b}}_3\mathrm{y}+{\mathrm{c}}_3=0$  are to be concurrent is  $\left|\begin{array}{ccc}{\mathrm{a}}_1& {\mathrm{b}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\\ {\mathrm{a}}_3& {\mathrm{b}}_3& {\mathrm{c}}_3\end{array}\right|=0$
Hence,  
$\left|\begin{array}{ccc}2& 1& -\mathrm{k}\\ 1& -1& 1\\ -2& -3& 8\end{array}\right|=0$  
Expand the determinant
 $\begin{array}{c}2\left(-8+3\right)-1\left(8+2\right)-\mathrm{k}\left(-3-2\right)=0\\ -10-10+5\mathrm{k}=0\\ \mathrm{k}=\frac{20}{5}\\ =4\end{array}$
Therefore, $k=\boxed{4}$