The distance of the point (5,3,-1) from the plane passing through points (2,1,0),(3,-2,4) and (1,-3,3) is

Step-by-step solution

The plane containing points A=(2,1,0) , B=(3,-2,4) , and C=(1,-3,3) has the vectors AB = (1, -3, 4) and AC = (-1, -4, 3) . A normal vector to the plane is given by their cross product: n = AB AC = 7 i - 7 j - 7 k . This simplifies to n = (1, -1, -1) by scaling. Using the point-normal form with A=(2,1,0) and normal (1,-1,-1) : 1(x-2) - 1(y-1) - 1(z-0) = 0 Simplifying yields the plane equation x - y - z - 1 = 0 . The distance from point (5,3,-1) to this plane is calculated as: D = |1(5) + (-1)(3) + (-1)(-1) + (-1)| 1