Math plumb line method?

You should first calculate the Hessian normal form. This allows you to calculate the distance of point P from the plane. Once you know this distance, you can add the normalized normal vector to point P, stretched by the changed sign, to obtain the perpendicular foot. Generally, it looks like this (vectors are in bold):

Step 1: Calculate HNF (here E is given in parametric form)

E: x = p + r u + s v

1.1. Calculate the normal vector

n = u × v

1.2. Normalize the normal vector

n = 1/| n | n

1.3. Setting up the HNF

E: ň • ( x – p ) = 0

E: ň • x – k = 0

(with k = ň • p )

2nd step: Calculate the distance from the point to the plane (I call the point whose perpendicular foot point F we want to know Q )

ň • Q – k = φ

|φ| is the distance between Q and E.

Step 3: Calculate the plumb line base point

F = Q – φ ň

4th step (optional): To check, you can check whether F is even in the plane.

ň • F – k ≟ 0

Given your circumstances, it would look like this:

Step 1: Calculate HNF, e.g.

E: x = (2, 0, 2) + r (1, 1, 2) + s (2, 3, 5)

n = (1, 1, 2) × (2, 3, 5) = (–1, –1, 1)

| n | = √3 => ň = 1/√3 (–1, –1, 1)

E: ň • ( x – p ) = 0

E: 1/√3 (–1, –1, 1) • ( x – (2, 0, 2)) = 0

E: 1/√3 (–1, –1, 1) • x = 0

Here k = 0, so the plane passes through the origin (0, 0, 0).

Step 2: Calculate distance from point plane

1/√3 (–1, –1, 1) • (5, –7, –8) = –2√3

Step 3: Calculate the plumb line base point

F = P – (–2√3) ň

F = (5, –7, –8) – (–2√3) 1/√3 (–1, –1, 1)

F = (3, –9, –6)

4th step (optional): To check, you can check whether F is even in the plane.

1/√3 (–1, –1, 1) • (3, –9, –6) = 0

Actually true.