The lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, if $k$ equals

The given lines are $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$

The condition that the lines are to be coplanar 

$\left|\begin{array}{ccc}2& k+1& -1\\ 2& 3& 4\\ 1& 2& 1\end{array}\right|=0$

It implies that 

   $\begin{array}{rcl}2\left(3-8\right)-\left(k+1\right)\left(2-4\right)-1\left(4-3\right)& =& 0\\ -10+2k+2-1& =& 0\\ 2k-9& =& 0\\ k& =& \frac{9}{2}\end{array}$

Therefore, $k=\boxed{\frac{9}{2}}$