Hello student,
First of all our aim shall be to find out the equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6).
The equation of the plane determined by the points (x
1, y
1, z
1), (x
2, y
2, z
2) and (x
3, y
3, z
3) is given by
So,
Hence, this gives (x-3)12 – (y+1)16 + (z-2)12 = 0
Hence, the eqaution of the plane is 12x-16y+12z = 76.
or, 3x – 4y + 3z = 19.
Now, the distance of the plane 3x – 4y + 3z = 19 from the point (6, 5, 9) is
=
