Find the coordinates of the point where the line through $(3, - 4, - 5)$ and $(2, - 3,1)$ crosses the plane$2x + y + z = 7$ .

The equation of the line through $(3, - 4, - 5)$ and $(2, - 3,1)$ is

$\cfrac{{x - 3}}{{2 - 3}} = \cfrac{{y + 4}}{{ - 3 + 4}} = \cfrac{{z + 5}}{{1 + 5}}$

i.e., $\cfrac{{x - 3}}{{ - 1}} = \cfrac{{y + 4}}{1} = \cfrac{{z + 5}}{6}$ $= k$ ...(1)
Any point on (1) is $(3 - k, - 4 + k, - 5 + 6k)$ ….(2)

This point lies on $2x + y + z = 7$ .

$\therefore$ $2(3 - k) + ( - 4 + k) + ( - 5 + 6k) = 7$

$\Rightarrow$ $6 - 2k - 4 + k - 5 + 6k = 7 \Rightarrow 5k = 10 \Rightarrow k = 2$

Putting in (2), the required point is $(3 - 2, - 4 + 2, - 5 + 12)$ i.e, $(1, - 2,7)$