The triangles in question, of area 5, 10 and 17, are proportional to the altitude from the point to the respective sides or sides extended, as the bases of the triangles formed are all the same length. If we make a hexagon with 5, 10 and 17 as the altitudes, the areas will be too big by a factor that is one half of one side of this intermediate hexagon.

First, to construct this intermediate hexagon with triangles' altitudes being 5, 10 and 17:

I constructed that hexagon using Geometer's Sketchpad with the following procedure:

Construct a vertical line segment of length 5 cm.

Then a line segment of length 10 cm at a 120° angle to the first, at point Q.

Then a line segment of length 17 cm at a 120° angle to each of the other two segments again meeting at Q.

Lines perpendicular to each of these segments were constructed at each one's end that was not Q. Their intersections formed the vertices of a large equilateral triangle and if each side is divided into three equal segments the middle segment of each forms a side of the hexagon in question.

The measured length of this in GSP is 12.31681 cm. When this is divided by sqrt(3) -- a likely irrational factor -- the quotient is  7.11111.... Assuming this is 7 + 1/9 (an assumption because of the limited precision of what GSP presents), one side of the hexagon is (7+1/9)*sqrt(3) or 64*sqrt(3)/9 ~= 12.316805742712.

But, as mentioned, the area is bloated by half this value as a factor. To get the area (and its subdivisions) down to the sought values we need to divide the linear sides by the square root of this halved value.

Half of 64*sqrt(3)/9 is 32*sqrt(3)/9. The square root of that value (to get a reduction factor for sides as opposed to area) is 4*sqrt(2)/3^(3/4).

So we want 64*sqrt(3)/9 / (4*sqrt(2)/3^(3/4)) = 8*sqrt(2)/3^(3/4) ~= 4.963225915211.