the number of solutions (x,y,z) to the system of equations

x+2y+4z = 9

4yz+2xz+xy= 13

xyz = 3 such that at least 2 of (x,y,z) are integers is??? pls answer quickly

Let the roots of the system equation are : 
α = x , β = 2y and γ = 4z. Thus :
α + β + γ = x + 2y + 4z = 9
αβ + βγ + γα = 2xy + 8yz + yzx = 2(4yz + 2xz + xy) = 26
αβγ = 8xyz = 24

Thus, our polynomial should be :
P³ – 9P + 26P – 24 = 0,
(P – 2)(P – 3)(P – 4) = 0

Since our roots are α = x , β = 2y and γ = 4z, so :
(x, 2y, 4z) = (2, 3, 4) or its permutations, or 6 combinations.
However, note that one case if x = 4, 2y = 3, and 4z = 2,
(x, y, z) = (4, 3/2, 1/2) which two of the roots are not an integer.

Excluding of this case, we have five solutions.