Find the equation of the plane determined by the points A(3, -1, 2), B(5, 2, 4) and C(-1, -1, 6) and hence find the distance between the plane and the point P(6, 5, 9)

The equation of the plane through three non-collinear points A(3, -1, 2), B(5, 2, 4) and C(-1, -1, 6) can be expressed as
|x -3 y+1 z-2|
|5-3 2+1 4-2| = 0
|-1-3 -1+1 6-2|

|x-3 y+1 z-2|
|2 3 2| = 0
|-4 0 4|
Expanding the determinant, we get:
(12-0)(x-3) - (8-8)(y+1) + (0+12)(z-2) = 0
12(x-3) + 12(z-2) = 0
12x - 36 + 12z - 24 = 0
12x + 12z -60 = 0
12x + 12z = 60
x + z = 5
is the required equation.

Now, distance of P(6, 5, 9) from the plane x + z = 5 is given by
|6 + 9 - 5| / √(1² + 0² + 1²) = |10| / √2 = 5√2 sq.units.