Let $\omega$ be the unit circle in the $xy$-plane in $3$-dimensional space. Find all points $P$ not on the $xy$-plane that satisfy the following condition: There exist points $A,B,C$ on $\omega$ such that
$$ \angle APB = \angle APC = \angle BPC = 90^\circ.$$
1. **Understanding the Problem:**
   We need to find all points \( P \) not on the \( xy \)-plane such that there exist points \( A, B, C \) on the unit circle \( \omega \) in the \( xy \)-plane satisfying \( \angle APB = \angle APC = \angle BPC = 90^\circ \).

2. **Geometric Interpretation:**
   Consider the unit circle \( \omega \) in the \( xy \)-plane. Let \( A, B, C \) be points on \( \omega \). The condition \( \angle APB = \angle APC = \angle BPC = 90^\circ \) implies that \( P \) is the orthocenter of the triangle \( \triangle ABC \) when projected onto the \( xy \)-plane.

3. **Projection onto the \( xy \)-plane:**
   Let \( H \) be the orthocenter of \( \triangle ABC \). The orthocenter \( H \) lies in the \( xy \)-plane. The point \( P \) must be such that its projection onto the \( xy \)-plane is \( H \).

4. **Distance from the \( xy \)-plane:**
   Let \( P = (x, y, z) \). The projection of \( P \) onto the \( xy \)-plane is \( (x, y, 0) \). Since \( H \) is the orthocenter of \( \triangle ABC \), we have \( (x, y, 0) = H \).

5. **Power of a Point:**
   The power of the orthocenter \( H \) with respect to the unit circle \( \omega \) is given by:
   \[
   \text{Pow}(H) = OH^2 - 1
   \]
   where \( O \) is the origin. Since \( H \) lies on the \( xy \)-plane, \( OH = \sqrt{x^2 + y^2} \). Therefore,
   \[
   \text{Pow}(H) = x^2 + y^2 - 1
   \]

6. **Distance Calculation:**
   The distance \( z \) from \( P \) to the \( xy \)-plane is given by:
   \[
   z = \sqrt{\frac{\text{Pow}(H)}{2}} = \sqrt{\frac{x^2 + y^2 - 1}{2}}
   \]

7. **Equation for \( P \):**
   Since \( P \) must satisfy the condition \( x^2 + y^2 + 2z^2 = 1 \), we substitute \( z \):
   \[
   x^2 + y^2 + 2 \left( \sqrt{\frac{x^2 + y^2 - 1}{2}} \right)^2 = 1
   \]
   Simplifying, we get:
   \[
   x^2 + y^2 + (x^2 + y^2 - 1) = 1
   \]
   \[
   2(x^2 + y^2) - 1 = 1
   \]
   \[
   2(x^2 + y^2) = 2
   \]
   \[
   x^2 + y^2 = 1
   \]

8. **Final Condition:**
   Therefore, the points \( P \) must satisfy:
   \[
   x^2 + y^2 + 2z^2 = 1
   \]

The final answer is \(\boxed{x^2 + y^2 + 2z^2 = 1}\).