Q.If lines $\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}$ and $\dfrac{x-3}{1} = \dfrac{y-k}{2} = \dfrac{z}{1}$ intersect each other then find $k$.

[!TLDR] Impose $(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=0$ with the unknown $k$ in $\vec a_2$. …

The lines are $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}$, through $A(1,-1,1)$ with direction $(2,3,4)$, and $\dfrac{x-3}{1}=\dfrac{y-k}{2}=\dfrac{z}{1}$, through $B(3,k,0)$ with direction $(1,2,1)$.

$\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\2&3&4\\1&2&1\end{vmatrix}=\hat i(3(1)-4(2))-\hat j(2(1)-4(1))+\hat k(2(2)-3(1))=-5\hat i+2\hat j+\hat k$ …

Read off points and direction ratios (one point has the unknown $k$), form $\vec a_2-\vec a_1$ symbolically, apply the intersection c …

Substituting $k$ into the wrong coordinate slot; sign error turning $y-k$ into a point coordinate (the point's $y$-value is $+k$, since the line is written as $\ …