Determinant form of plane through point and line

A plane through a given line is a member of the pencil of planes through that line; making it pass through the given point fixes the member.

The line as two planes

x = py + q = rz + s means x − (py + q) = 0 and x − (rz + s) = 0. Write

u = x, v = py + q, w = rz + s.

The line is u = v = w.

Plane through the line

Any plane through the line is (u − v) + λ(u − w) = 0, i.e.

(1 + λ)u − v − λw = 0.

Its coefficients of u, v, w are (1 + λ, −1, −λ), and they add up to 0. It is a plane because u, v, w are linear in x, y, z.

The determinant

Let u₀ = α, v₀ = pβ + q, w₀ = rγ + s for the point (α, β, γ). Expand

| u   v   w  |
| u₀  v₀  w₀ |  = u(v₀ − w₀) − v(u₀ − w₀) + w(u₀ − v₀) = 0
| 1   1   1  |

Properties of this equation:

1. It is linear in u, v, w, so it represents a plane.
2. The coefficients (v₀ − w₀), −(u₀ − w₀), (u₀ − v₀) add up to 0, like the pencil above, so it belongs to the pencil of planes through the line (put u = v = w = t: the first and third rows become proportional, so the determinant is zero for every point of the line).
3. Put (u, v, w) = (u₀, v₀, w₀): rows 1 and 2 are equal, so the determinant is 0. The point (α, β, γ) lies on the plane.

A plane through the line and the point is unique, hence the plane required is

| x py+q rz+s ; α pβ+q rγ+s ; 1 1 1 | = 0.