Sure thing... If I understand the problem correctly, it doesn't really matter what the exact distributions of x, y, and z are. It only matters that they are identically distributed. Also, note that the condition x+y+z = 1 is unchanged by permuting the letters x,y,z. This, together with their being identically distributed means that
E{x|x+y+z=1} = E{y|x+y+z=1} = E{z|x+y+z=1}
Also, think about what E{x+y+z|x+y+z=1} would be, and remember the additive property of the expected value.
edit: think simple. no hard integrals needed, which was the first thing that came to my mind when I read the problem.