Given line is
$$
\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda
$$

So, $(x, y, z)$ is $(2 \lambda+1,3 \lambda-1,4 \lambda+1)$ and this point is lies on given line. This point also lies on line.
$$
\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}
$$

So, this point satisfies equation
$$
\begin{array}{rlrl}
& & \frac{2 \lambda+1-3}{1} & =\frac{3 \lambda-1-k}{2}=\frac{4 \lambda+1}{1} \\
\Rightarrow & & \frac{2 \lambda-2}{1} & =\frac{4 \lambda+1}{1} \\
\Rightarrow & 2 \lambda-2 & =4 \lambda+1 \\
\Rightarrow & & 2 \lambda & =-3 \\
\Rightarrow & & \lambda & =-\frac{3}{2}
\end{array}
$$

Now, $\frac{3 \lambda-1-k}{2}=\frac{2 \lambda-2}{1}$
$$
\Rightarrow \frac{3 \times-\frac{3}{2}-1-k}{2}=\frac{2 \times-\frac{3}{2}-2}{1}
$$
$\begin{array}{ll}\Rightarrow & \frac{-\frac{9}{2}-1-k}{2}=\frac{-3-2}{1} \Rightarrow-\frac{11}{2}-k=-10 \\ \Rightarrow & -\frac{11}{2}+10=k \Rightarrow k=\frac{9}{2}\end{array}$