Let SABC be a regular triangular pyramid. Find the set of all points $ D (D! \equal{} S)$ in the space satisfing the equation $ |cos ASD \minus{} 2cosBSD \minus{} 2 cos CSD| \equal{} 3$.
1. **Understanding the Geometry**:
   - We have a regular triangular pyramid \( SABC \) where \( SA = SB = SC \) and \( AB = BC = CA \).
   - We need to find the locus of points \( D \) such that \( D \neq S \) and the equation \( | \cos \delta_A - 2 \cos \delta_B - 2 \cos \delta_C | = 3 \) holds, where \( \delta_X = \angle XSD \) for \( X \in \{A, B, C\} \).

2. **Using Symmetry**:
   - Since \( SABC \) is a regular triangular pyramid, it is symmetric with respect to the plane passing through \( S \) and the centroid of \( \triangle ABC \).
   - Let \( G \) be the centroid of \( \triangle ABC \). The symmetry implies that any point \( D \) satisfying the given condition must lie on the plane passing through \( S \) and perpendicular to the plane \( ABC \).

3. **Analyzing the Equation**:
   - The equation \( | \cos \delta_A - 2 \cos \delta_B - 2 \cos \delta_C | = 3 \) must be satisfied.
   - Consider the angles \( \delta_A, \delta_B, \delta_C \) formed by \( D \) with \( S \) and the vertices \( A, B, C \).

4. **Using Trigonometric Identities**:
   - We know that for any point \( D \) in space, the sum of the cosines of the angles formed with respect to a fixed point \( S \) and the vertices of a regular triangle is constant.
   - Specifically, for a regular triangular pyramid, the sum of the cosines of the angles formed by \( D \) with \( S \) and the vertices \( A, B, C \) is related to the geometry of the pyramid.

5. **Solving the Equation**:
   - We need to solve \( | \cos \delta_A - 2 \cos \delta_B - 2 \cos \delta_C | = 3 \).
   - Given the symmetry and the regularity of the pyramid, we can infer that the angles \( \delta_A, \delta_B, \delta_C \) must be such that the equation holds true.

6. **Conclusion**:
   - The only way for the equation \( | \cos \delta_A - 2 \cos \delta_B - 2 \cos \delta_C | = 3 \) to hold is if \( D \) lies on the plane passing through \( S \) and perpendicular to the plane \( ABC \).
   - This plane is the locus of all points \( D \) satisfying the given condition.

The final answer is the plane passing through \( \boxed{ S } \) and perpendicular to the plane \( ABC \).