The straight lines (x-2)/1=(y-3)/1=(z-4)/(-k) and (x-1)/k=(y-4)/2=(z-…

Any point on the first line can be takes as

$P_1=\left(r_1+2, r_1+3,-k r_1+4\right)$

These lines will intersect if for some r₁ and r₂ we have

$r_1+2=kr_2+1,$

$r_1+3=2 r_2+4,$

$-kr_1+4=r_2+5$

∴  $r_1-kr_2+1=0, r_1=2 r_2+1 $

$\Rightarrow r_2=\frac{2}{k-2}, r_1=\frac{k+2}{k-2}$

putting these values in the last condition, we get

$k^2+3 k=0$

⇒ k = {-3, 0}