Pair the factors of N such that each pair's product equals N; if N is a perfect square then sqrt(N) will be by itself. 

The geometric mean of each pair is sqrt(N), so then collectively the geometric mean of all the factors of N is sqrt(N).

But then this makes the product of all factors easy to compute as N^(F/2).

Lets go a bit deeper and let N be a number with exactly three prime factors of 2, 3, and 5. Then N = 2^x * 3^y * 5^z and F = (x+1)*(y+1)*(z+1).

Then N^(F/2) = 2^[x*(x+1)*(y+1)*(z+1)/2] * 3^[(x+1)*y*(y+1)*(z+1)/2] * 5^[(x+1)*(y+1)*z*(z+1)/2].

Now to tune in on this problem, N^(F/2) is given as 6^12 * 10^24 = 2^36 * 3^12 * 5^24.

Then we can create the following system by equating the two expressions for N^(F/2) and matching the exponents

x*(x+1)*(y+1)*(z+1)/2 = 36

(x+1)*y*(y+1)*(z+1)/2 = 12

(x+1)*(y+1)*z*(z+1)/2 = 24

Taking ratios of these equations gives us that x:y:z = 3:1:2.  Since we are interested in positive integers then the obvious answer is x=3, y=1, and z=2, which does satisfy the system of equations.  Then N = 2^3 * 3^1 * 5^2 = 600.