You actually don't have enough information to unambiguously define the surface.
Here's the classic illustration of that fact. Let's say we have these three projections:
[xs, ys, zs] = sphere;
grey = [.75 .75 .75];
h(1) = surface(xs,ys,zeros(size(zs))-1.5);
h(2) = surface(zeros(size(xs))+1.5,ys,zs);
h(3) = surface(xs,zeros(size(ys))+1.5,zs);
set(h,'FaceColor',grey,'EdgeColor','none','FaceLighting','none');
axis vis3d
box on
xlim([-1.5 1.5])
ylim([-1.5 1.5])
zlim([-1.5 1.5])
view(3)
What's the surface that's defined by those? It's obviously this sphere, isn't it?
hold on
hs = surf(xs,ys,zs,ones(size(zs)),'EdgeColor','none','FaceLighting','gouraud');
hold off
axis vis3d
camlight
Not really. There are other possibilities. Try this one:
delete(hs)
[y,x,z] = ndgrid(linspace(-1,1,100));
v = max(sqrt(x.^2+y.^2),max(sqrt(y.^2+z.^2), sqrt(z.^2+x.^2)));
isosurface(x,y,z,v,1)
Try rotating that around. You'll see that it lines up with the 3 projections.
So the point of that long winded example is that there are several possible surfaces which can yield the same projections. Going backwards isn't simple. In fact it's a particularly rich field of computer graphics.
The way it is usually done for boats is to have a series of "stations" and "fair" them. You can find a description of that process here, but you don't have enough information to do that. You can probably come up with an approximation that looks good. It's going to involve things like choosing the first partial derivatives along the keel (how "sharp" the bottom of the boat is).
