Find the coordinates of the point where the line through `(3,-4,-5)` and (2,-3,1) crosses the plane passing through the points (2,2,1),(3,0,1) and (4,-1,0).

Explanation

We know that the equation of a plane passing throgh the points `A(x_(1),y_(1),z_(1)), B(x_(2),y_(2),z_(2))` and `C(x_(3),y_(3),z_(3))` is given by
`|{:(x-x_(1),y-y_(1),z-z_(1)),(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(x_(3)-x_(1),y_(3)-y_(1),z_(3)-z_(1)):}|=0`.
`therefore` the equation of the plane passing through the points A(2,2,1), B(3,0,1) and C(4,01,0) is given by
`rArr |{:(x-2,y-2,z-1),(3-2,0-2,1-1),(4-2,-1-2,0-1):}|=0 rArr |{:(x-2,y-2,z-1),(1-2,0),(2,-3,-1):}|=0`
`rArr (x-2)(2-0)-(y-2)(-1-0)+(z-1)(-3+4)=0`
`rArr 2(x-2)+1.(y-2)+1.(z-1)=0 rArr 2x+y+z=7`............(i)
The equation of the line passing through the points P(3,-4,-5) and Q(2,-3,1) is given by
`(x-3)/(2-3)=(y+4)/(-3+4)=(z+5)/(1+5) rArr (x-3)/(-1)=(y+4)/1=(z+5)/6= lambda` (say).
Any general point on line PQ is given as `(-lambda+3,lambda-4,6lambda-5)`.
For some value of `lambda`, let the point `R(-lambda+3,lambda-4,6lambda-5)`.
lie on the plane (i). Then,
`2(lambda+3)+(lambda-4)+(6lambda-5)=7 rArr 5lambda=10 rArr lambda-2`.
Putting, `lambda=2`, we get the point of intersection of the given line and the given plane as `R(-2+3,2-4,12-5),`i.e., `R(1,-2,7)`.