Question

Find the 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 the point P(6, 5, 9) from the plane.

Answer 1

The equation of the plane through three non-collinear points A(3, –1, 2), B(5, 2, 4) and (–1, –1, 6) can be expressed as

=>12(x - 3) - 16(y + 1) + 12(z - 2) = 0
=>12x - 16y + 12z - 76 = 0 => 3x - 4y + 3z - 19 = 0 is the required equation.

Now distance of P(6, 5, 9) from the plane is given by