6 On the surface of a sphere with radius 1, there are four non-coplanar points $A, B, C, D$, and $AB=CD=x, BC=DA=y, CA=BD=z$. Then $x^{2}+y^{2}+z^{2}$ equals ( ).
(A) 2
(B) 4
(C) 8
(D) 16
$6 \mathrm{C}$ Hint: Construct a rectangular parallelepiped such that the six edges of $A B C D$ are the diagonals of one of its faces. At this time, the length of the space diagonal of the rectangular parallelepiped is $\sqrt{\left(x^{2}+y^{2}+z^{2}\right) / 2}$, which is exactly equal to the diameter of the circumscribed sphere, so $x^{2}+y^{2}+z^{2}=8$.