Find the distance between the point P (6, 5, 9) and the plane determined by points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6). ?

Hello student,
Please find the answer to your question below
The plane determined by points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6) is

Which gives 3x-4y+3z=19
distance between the point P (6, 5, 9) and the plane3x-4y+3z=19 is
distance=(3(6)-4(5)+3(9)-19)/sqrt(9+16+9)
=6/units