If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$…
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
$-1$
$\frac{2}{9}$
$\frac{9}{2}$
$0$
Solution
Any point on $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=t$ is $(2 t+1,3 t-1,4 t+1)$
And any point on $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mathrm{s}$ is $(s+3,2 s+k, s)$
Given lines are intersecting
$\Rightarrow \mathrm{t}=-\frac{3}{2} \text { and } \mathrm{s}=-5 \quad \therefore \mathrm{k}=\frac{9}{2}$