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 $(2t+1,3t-1,4t+1)$.Any point on the second line is $(s+3,2s+k,s)$.At intersection:$2t+1=s+3⟹2t-s=2$ ...(1)$4t+1=s⟹s-4t=1$ ...(2)Adding (1) and (2):$-2t=3⟹t=-3/2$.Then $s=4(-3/2)+1=-6+1=-5$.Now use the $y$ coordinate:$3t-1=2s+k$$3(-3/2)-1=2(-5)+k$$-4.5-1=-10+k$$-5.5=-10+k⟹k=10-5.5=4.5=9/2$.