surface area

a vertex of the unit cube is said to be visible from P if it can be connected to P by a straight line that does not pass through the cube.

from any point P outside a unit cube, 4,6, or 7 vertices are visible in the same sense as in the case of the square. Connecting point P to each of these vertices gives 1,2, or 3 square-based pyramids, which make up the visible volume of P. determine the surface area of set of all points P for which the visible volume is 20, and is a polyhedron.

In the space around the cube that sees one side of the cube, a point P is h=3*V/B=60 from the cube side and the area of these are 6*1*1=6
In the space where two sides are visible, the volume is B(x+y)/3 where (x,y) is the displacement from the visible edge which connects the two sides, in the plane normal to both of them. These points forms sides that connects each two of the 6 sides of points that sees one side, and should be of area 12*1*60*sqrt(2).
In the space around the cube that sees 3 sides, the volume is B(x+y+z)/3 where (x,y,z) is the displacement from the visible vertex that connects the 3 sides. These points forms triangular sides that connects each 3 adjacent of the 12 sides of points that sees two sides, their area should be 8*60*sqrt(2)*60*sqrt(2)/sqrt(3)
The polyhedron with a total of 26 sides should have the surface area of 6+720*sqrt(2)+57600/sqrt(3) which is about 34279.6