If the lines \(\frac{x-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-1}{4}\) and \(\frac{x-3}{1}=\frac{\mathrm{y}-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}\) interect, then the value of the \(\mathrm{k}\) is

Let \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right) \equiv(1,-1,1)\) and \(\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right) \equiv(3, \mathrm{k}, 0)\)
Let \(\left(\mathrm{a}_{1}, \mathrm{~b}_{1}, \mathrm{c}_{1}\right) \equiv(2,3,4)\) and let \(\left(\mathrm{a}_{2}, \mathrm{~b}_{2}, \mathrm{c}_{2}\right) \equiv(1,2,1)\)
Since, the lines intersect
\(\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|=0\)
\(\left|\begin{array}{ccc}2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1\end{array}\right|=0\)
\(2(3-8)-(k+1)(2-4)-1(4-3)=0\)
\(-10+2 k+2-1=0 \Rightarrow k=\frac{9}{2}\)