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

$\frac{9}{2}$

We know that the lies

$\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$

intersect, if

$\begin{vmatrix}x_2-x_1 & y_2-y_1 & z_2-z_1\\l_1 & m_1 & n_1\\l_2 & m_2 & n_2\end{vmatrix}=0$

So, the given lines will intersect, if

$\begin{vmatrix}3-1 & k+1 & 0-1\\2& 3& 4\\l& 2 & 1\end{vmatrix}=0$

$⇒2(3-8) - (k+1)(2-4) - (4-3) = 0 ⇒k = \frac{9}{2}$