Three Dimensional Geometry — Class 12 Maths Solution

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

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

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

This lies on ZX plane $(y = 0)$
$\therefore$ $1 + 3k = 0 \Rightarrow k = - \cfrac{1}{3}$

Putting in (2), $\left( {5 + \cfrac{2}{3},1 - 1,6 + \cfrac{5}{3}} \right)$ i.e., $\left( {\cfrac{{17}}{3},0,\cfrac{{23}}{3}} \right)$

which is the required point.