Find the equation of a plane passing through the points A(2,1,2) and B(4,-2,1) and perpendicular to plane `vec(r ).(hat(i)-2hat(k))=5`. Also, find the coordinates of the point, where the line passing through the points (3,4,1) and (5,1,6) crosses the plane thus obtained.

Explanation

Let P(x,y,z) be any point on the plane in which A(2,1,2) and B(4,-2,1) lie.
`:. Vec(AP)` and `vec(AB)` lie on required plane.
Also required plane is perpendicular to given plane `vecr. (hati-2hatk) = 5`
`:.` normal to given plane `vecn_(1) = (hati -2hatk)` lie on reqired plane.
`rArr vec(AP), vec(AB)` and `vecn_(1)` are coplanar,
Where `vec(AP) = (x-2)hati + (y-1) + (z-2)hatk`
`vec(AB) = 2hati - 3hatj - hatk`
`rArr` Scalar triple product `[vec(AP) vec(AB) vec(n_(1))] = 0`
`rArr |{:(x-2,y-1,z-2),(2,-3,-1),(1,0,-2):}|= 0`
`rArr (x-2) (6-0)-(y-1)(-4+1)+(z-2)(0+3)=0`
`rArr 6x-12+3y-3+3z-6=0`
`rArr 2x+y+z = 7 "............."(1)`
Line passing through points `L(3,4,1)` and `M(5,1,6)` is
`rArr (x-3)/(2)=(y-4)/(-3)=(z-1)/(5)=lambda"............"(2)`
General point on the line is `Q(2lambda + 3, -3lambda+4,5lambda+1)`
As line (2) crosses plane (1) so point Q should satisfy equation (1)
`:. 2(2lambda + 3)+ (-3lambda+4) + (5lambda+1) = 7`
`4lambda+6-3lambda+4+5lambda+1=7`
`6lambda = - 4`
`lambda = -2/3`
`Q (-4/3 + 3,2+4,-(10)/(3)+1) = Q(5/3,6,-7/3)`