Some preliminary results
Lemma 1
If A, B, C are non-collinear p-points, then there is a plane Π cutting the rays
at A', B', C', respectively, such that ΔA'B'C' is acute-angled.
Proof
Choose unit vectors so that A = [a], B = [b], C = [c].
Observe that vectors u,v make angle θ where cos(θ) = u.v/|u||v|, so make
an acute angle if and only if u.v > 0.
Replacing a by -a if necessary, we may assume that a.c ≥ 0.
Replacing b by -b if necessary, we may assume that b.(a+c) ≤ 0.
We choose A', B', C' with position vectors a, λb, c, with λ to be chosen later.
Angle A'B'C' is acute if and only if (a-λb).(c-λb) > 0,
i.e. a.c -λ(a+c).b+λ2>0.
Angle B'A'C' is acute if and only if (λb-a).(c-a) > 0,
i.e. λb.(c-a)-a.c+1>0.
Angle A'C'B' is acute if and only if (λb-c).(a-c) > 0, i.e. λb.(a-c)-a.c+1>0.
The first is true for all positive λ since a.c ≥ 0, b.(a+c) ≤ 0.
Since a, c are non-parallel, a.c<1, so that the second and third will be true
for sufficiently small λ.
Lemma 2
If ΔPQR is acute angled, then we can embed it in R3 with P, Q, R on
the positive x-, y-, and z-axes respectively.
Proof
Let |PQ| = r, |QR| = p, |RP| = q.
As ΔPQR is acute angled, the Cosine Rule shows that
p2+q2-r2,
p2+r2-q2
and q2+r2-p2 are positive.
We can embed the triangle as required if we can find points (x,0,0), (0,y,0)
and (0,0,z) such that x2+y2=r2, y2+z2=p2 and z2+x2=q2.
These have solution
x2=(q2+r2-p2)/2,
y2=(p2+r2-q2)/2,
z2=(p2+q2-r2)/2.
Since the values are positive, we can take x, y, z as the positive roots.
Lemma 3
For λ ≠ 0, the following projective transformations correspond to perspectivies.
r([x,y,z]) = [x/λ,y,z],
s([x,y,z]) = [x,y/λ,z],
t([x,y,z]) = [x,y,z/λ].
Proof
We consider t, the others are similar.
Consider the perspectivity mapping Π : z=1 to Π' : z=λ with centre O.
We choose axes on both planes lying over the x- and y-axes.
Now, (x,y,1) on Π maps to (λx,λy,λ) on Π', and this projects to (λx,λy,1) on Π.
As a projective transformation, this maps [x,y,1] to [λx,λy,1] = [x,y,1/λ],
so is equal to t.