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

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)`