This can basicly be reduced to a plane geometry problem. The vertex $D$ is projected onto the orthocenter $H$ of $ABC$ (this follows from the conditions $AB\perp CD,\ AC\perp BD,\ AD\perp BC$). This means that the midpoints $U,V,T$ of $DA,DB,DC$ are projected onto the midpoints $X,Y,Z$ of $HA,HB,HC$. Let $M,N,P$ be the midpoints of $BC,CA,AB$. The points $M,N,P,X,Y,Z$ liue on the nine-point circle of $ABC$, so the points $U,V,T$ lie on a circle of radius equal to $\frac R2$ ($R$ is the circumradius of $ABC$) which lies on a plane parallel to $ABC$ and which has its center directly above the nine-point center of $ABC$. It's is now clear that $M,N,P,U,V,T$ lie on a sphere which has its center in the midpoint of the segment formed by the centers of $(MNP),(UVT)$.