We consider a fixed point $P$ in the interior of a fixed sphere$.$ We construct three segments $PA, PB,PC$, perpendicular two by two$,$ with the vertexes $A, B, C$ on the sphere$.$ We consider the vertex $Q$ which is opposite to $P$ in the parallelepiped (with right angles) with $PA, PB, PC$ as edges$.$ Find the locus of the point $Q$ when $A, B, C$ take all the positions compatible with our problem.
1. **Define the problem setup:**
   - Let $P$ be a fixed point inside a sphere.
   - Construct three segments $PA, PB, PC$ such that they are mutually perpendicular and the points $A, B, C$ lie on the sphere.
   - Construct a parallelepiped with $PA, PB, PC$ as edges, and let $Q$ be the vertex opposite to $P$.

2. **Understand the geometric configuration:**
   - Since $PA, PB, PC$ are mutually perpendicular, they form the edges of a rectangular parallelepiped.
   - The point $Q$ is determined by the vectors $\overrightarrow{PA}, \overrightarrow{PB}, \overrightarrow{PC}$.

3. **Express $Q$ in terms of $P, A, B, C$:**
   - Let $P$ have coordinates $(x_0, y_0, z_0)$.
   - Let $A, B, C$ have coordinates $(x_1, y_0, z_0)$, $(x_0, y_1, z_0)$, $(x_0, y_0, z_1)$ respectively.
   - Then, $Q$ will have coordinates $(x_1, y_1, z_1)$.

4. **Determine the locus of $Q$:**
   - Since $A, B, C$ lie on the sphere, their coordinates satisfy the equation of the sphere:
     \[
     (x_1 - x_0)^2 + (y_0 - y_0)^2 + (z_0 - z_0)^2 = R^2
     \]
     \[
     (x_0 - x_0)^2 + (y_1 - y_0)^2 + (z_0 - z_0)^2 = R^2
     \]
     \[
     (x_0 - x_0)^2 + (y_0 - y_0)^2 + (z_1 - z_0)^2 = R^2
     \]
   - Therefore, $x_1, y_1, z_1$ must satisfy:
     \[
     (x_1 - x_0)^2 + (y_1 - y_0)^2 + (z_1 - z_0)^2 = R^2
     \]

5. **Conclusion:**
   - The coordinates of $Q$ must satisfy the equation of the sphere centered at $P$ with radius $R$.
   - Therefore, the locus of $Q$ is a sphere centered at $P$ with the same radius as the original sphere.

The final answer is $\boxed{\text{a sphere centered at } P \text{ with radius } R}$.