Show that the points `(1,-1,3) and (3, 4, 3)` are equidistant from the plane `5x +2y + -7z + 8 = 0.`

Explanation

Given equation of the plane is `5x+2y-7z+8=0`.
Distance of given plane from the point (1, -1, 3)
`P_(1)=|(5(1)+2(-1)-7(3)+8)/(sqrt(5^(2)+2^(2)+(-7)^(2)))|`
`=|(5-2-21+8)/(sqrt(78))|`
`:.P_(1)=|(-10)/(sqrt(78))|=(10)/(sqrt(78))` ...(1)
Distance of the plane from the point (3,4,3)
`P_(2)=|(5(3)+2(4)+(-7)(3)+8)/(sqrt(5^(2)+2^(2)+(-7)^(2)))|`
`=|(15+8-21+8)/(sqrt(78))|`
`:.P_(2)=(10)/(sqrt(78))` ....(2)
From equation (1) and (2), `P_(1)=P_(2)`.
`:.` The given points are equidistant from the plane
`5x+2y-7z+8=0`. `" "` Hence Proved.