It is easy to see that, if the coordinates (x, y, z) are connected by any two linear equations, these equations can always be brought into the foregoing form, and hence that the two linear equations represent a line.
Secondly, taking for greater simplicity the point Q to be coincident with the origin, and α′, β′, γ′, p to be constant, then p is the perpendicular distance of a plane from the origin, and α′, β′, γ′ are the cosine-inclinations of this distance to the axes (α′² + β′² + γ′² = 1). P is any point in this plane, and taking its coordinates to be (x, y, z) then (ξ, η, ζ) are = (x, y, z), and the foregoing equation p = α′ξ + β′η + γ′ζ becomes
α′x + β′y + γ′z = p,
which is the equation of the plane in question.
If, more generally, Q is not coincident with the origin, then, taking its coordinates to be (a, b, c), and writing p1 instead of p, the equation is
α′ (x − a) + β′ (y − b) + γ′ (z − c) = p1;
and we thence have p1 = p − (aα′ + bβ′ + cγ′), which is an expression for the perpendicular distance of the point (a, b, c) from the plane in question.
It is obvious that any linear equation Ax + By + Cz + D = O between the coordinates can always be brought into the foregoing form, and hence that such an equation represents a plane.
Thirdly, supposing Q to be a fixed point, coordinates (a, b, c), and the distance QP = ρ, to be constant, say this is = d, then, as before, the values of ξ, η, ζ are x − a, y − b, z − c, and the equation ξ² + η² + ζ² = ρ² becomes
(x − a)² + (y − b)² + (z − c)² = d²,