Nope, the maximum area of a triangle with a perimeter of 12 is obviously an equilateral triangle of side 4 which has an area of 6.9, not enough. So we end up with only imaginary solutions.

Therefore the simplest convex solution must be a quadrilateral

Here is one way to make infinite parallelograms with the required perimeter and area:

Generate random side A for a rectangle of length between 2 and 4

generate the other side, B = 6 - A

calculate the area AB

remembering that the area of a parallelogram is sin(x)AB where x is the smaller of the two angles at the base we need

sin(x)AB = 8

x = asin(8/AB)

this gives x=90 for A=2, B=4

x = asin(8/9) for A=3, B=3

x = asin(8/8.75) for A=2.5, B=3.5

etc.