12 x 11 x 10 = 1320
The first can be any of 12
The second can be any of 11 (one already 1st)
The third can be any of 10 (2 already 1st and 2nd)

Therefore there are 12 x 11 x 10 permutations of this result = 1320.
This answer above is the answer I would normally have given. It is the equivalent of12P3 and equals 12!/9!.

HOWEVER as you see the answer has 4 digits and so I suspect that you want combinations rather than permutations. 12C3 = 12! /9!3! = 1320/6 = 220
This is the answer that you presumably want

It must be stated though that if you want combinations then you regard A finishing firrst B finishing 2nd and C finishing third as EXACTLY the same as C finishing first A finishing second and B finishing third or any of the other arrangements that you can make of these 3.
Three people can be arranged in 3! or 6 different ways
ABC, ACB, BAC, BCA, CAB, CBA
therefore for every combination of 3 people you can get 6 permutations. Therefore that is why the 1320 permutations is divided by 6 = 220 combinations.

IN conclusion – the answer that you probably want is 220 COMBINATIONS but I don’t like it. I would have thought 1320 PERMUTATIONS is betrter in this case because there IS a difference between first second and third places.
References :

Let O be the origin point of the two runners;
let A be the first runner;
let B be the second runner;
let α be the measure of the angle BÔD.

By definition of the problem:
OA = 3t is the distance made by the first runner.
OB = 4t is the distance made by the second runner.
α = 3π/4

The law of cosines state:
AB² = OA² + OB² – 2.OA.OB.cos(α)
AB² = (3t)² + (4t)² – 2.3t.4t.cos(3π/4)
AB² = 25t² – 24t².(-√2/2)
AB² = 25t² – 24t².(-√2/2)
AB² = t²(25 + 12√2)
AB = t.√(25 + 12√2)

So if AB, the distance between both runners, is 20 miles, elapsed time is:
t = AB / √(25 + 12√2) = 20 / √(25 + 12√2) ≈ 3,09 hours.

So the runners will lose contact approximatively 3 hours and 5 minutes after leaving each other.

The answer of antoine2546, that 10 hours, is both wrong and unrealistic since, in 10 hours, both runners would have run respectively 30 and 40 miles, way over the signal range of 20 miles!

Logically,
Dragon.Jade
References :

The distance ran by the north runner = 3t where t = time
The distance ran by the south east runner = 4t

The angle between their paths is constant = 90 + 45 = 135 degrees

let the direct distance between the runners be x(t).

Cosine rule says:

x(t)^2 = (3t)^2 + (4t)^2 – 2(3t)(4t)cos(135 deg)
x(t)^2 = 13t^2 – 24t^2cos(135) = t^2(13 – 24cos(135))

runners lose contact when x(t) > 20 miles ie
when
t^2(13 – 24cos(135)) > 400
t > sqrt(400/(13 – 24cos(135)))
cos(135) = -1/sqrt(2)
t > 13.3 hours (1 decimal)
References :

velocity = momentum/mass = 218kgm/s / 52 kg = 4.2 m/s
References :