Find the coordinates of the point where the line through $(5,1,6)$and $(3,4,1)$ crosses the XZ-plane.

The equation of line through $(5,1,6)$and $(3,4,1)$is

$\cfrac{{x - 5}}{{3 - 5}} = \cfrac{{y - 1}}{{4 - 1}} = \cfrac{{z - 6}}{{1 - 6}} \Rightarrow \cfrac{{x - 5}}{{ - 2}} = \cfrac{{y - 1}}{3} = \cfrac{{z - 6}}{{ - 5}} = k(say)$ …..(1)

Any point on (1) is$(5 - 2k,1 + 3k,6 - 5k)$ …..(2)

This lies on YZ-plane $(x = 0).$

$\therefore$ $5 - 2k = 0 \Rightarrow k = \cfrac{5}{2}$

Putting in (2), $\left( {5 - 5,1 + \cfrac{{15}}{2},6 - \cfrac{{25}}{2}} \right)$

i.e., $\left( {0,\cfrac{{17}}{2},\cfrac{{ - 13}}{2}} \right)$
which is the required point.