Find the point where the line passing through $(3,4,1)$ and $(5,1,6)$ intersects the $xy$-plane.
The direction vector the line is $\begin{pmatrix} 5 - 3 \\ 1 - 4 \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix},$ so the line is paramaterized by
\[\begin{pmatrix} 3 \\ 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 + 2t \\ 4 - 3t \\ 1 + 5t \end{pmatrix}.\]We want the $z$-coordinate to be 0, so $1 + 5t = 0.$  Then $t = -\frac{1}{5},$ so the point of intersection is $\boxed{\left( \frac{13}{5}, \frac{23}{5}, 0 \right)}.$
The direction vector the line is $\begin{pmatrix} 5 - 3 \\ 1 - 4 \\ 6 - 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix},$ so the line is paramaterized by
\[\begin{pmatrix} 3 \\ 4 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 + 2t \\ 4 - 3t \\ 1 + 5t \end{pmatrix}.\]We want the $z$-coordinate to be 0, so $1 + 5t = 0.$  Then $t = -\frac{1}{5},$ so the point of intersection is $\boxed{\left( \frac{13}{5}, \frac{23}{5}, 0 \right)}.$
The final answer is $\[ \boxed{\left( \frac{13}{5}, \frac{23}{5},$. I hope it is correct.