Find the vector equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6). Also find the distance of point P (6, 5, 9) from this plane.

The given points are A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6). Equation of any plane passing through A,

a(x-3)+b(y+1) + c(z-2) = 0

This plane passes through B and C as well so,

2a+3b+2c = 0 and

-4a+0b+4c =0

Solving these equations, we get,

Equation of the required plane,

3(x-3) -4(y+1) + 3(z-2) = 0

⇒ 3x – 4y + 3z – 19 =0 ….(1)

In vector form,

Now, distance of P(6, 5, 9) from (1),