If 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, then $k$  is equal to 

Any point on the first line is$\left(2r_1+1,3r_1-1,4r_1+1\right)$  and on the second line is  $\left(r_2+3,2r_2+k,r_2\right)$
The lines will intersect when 
$2r_1+1=r_2+3,3r_1-1=2r_2+k,4r_1+1=r_2$
$\Rightarrow 2r_1-r_2=2,4r_1-r_2=-1$
.$\Rightarrow r_1=-\frac{3}{2},r_2=-5$
and $k=3r_1-2r_2-1=-\frac{9}{2}+10-1=\frac{9}{2}.$