Hi Atom,
Your code doesn't work because if you multiply out X2^2 + Y2^2 + Z2^2, you don't get R^2. So that arc is not on the surface of the sphere. The correct arc can be gotten by taking the x,y,z column vector
R*[cos(T),sin(T),0]'
that does lie on the sphere and multiplying by an appropriate 3x3 rotation matrix to put that circular path into the plane x+y+z = 0. However, it can be seen that each of x,y,z is just a linear combination of cos(T) and sin(T), and with some fiddling around one can arrive at the following in place of what you have:
% don't need another hold on since the previous hold on still applies.
T = linspace(0,2*pi,100)
x = R*( cos(T)/sqrt(6) + sin(T)/sqrt(2));
y = R*( cos(T)/sqrt(6) - sin(T)/sqrt(2));
z = R*(-2*cos(T)/sqrt(6));
ind = z<0; % since it's a hemisphere, lop off the curve below z = 0
x(ind)=[]; y(ind)=[]; z(ind)=[];
plot3(x,y,z,'b','linewidth',5)
hold off
You could also restrict T to the appropriate range so that no z<0 points are created in the first place.
