Given a triangle $ABC$. Consider all the tetrahedrons $PABC$ with $PH$ -- the smallest of all tetrahedron's heights. 
Describe the set of all possible points $H$.
1. **Define the problem and notation:**
   - We are given a triangle \(ABC\) and need to consider all tetrahedrons \(PABC\) such that \(PH\) is the smallest of all tetrahedron's heights.
   - Let \(d(X, l)\) denote the distance from point \(X\) to line \(l\).

2. **Condition for the height \(PH\):**
   - By the problem's condition, the height from \(P\) to the plane \(ABC\) (denoted as \(PH\)) should be the smallest height among all possible tetrahedrons \(PABC\).
   - This implies that the height from \(A\) to the plane \(PBC\) should not be shorter than \(PH\). Mathematically, this is equivalent to saying that the area of \(\triangle ABC\) should be greater than or equal to the area of \(\triangle PBC\).

3. **Relate areas to distances:**
   - The area condition \([ABC] \ge [PBC]\) holds if and only if the perpendicular distance from \(A\) to line \(BC\) (denoted as \(d(A, BC)\)) is greater than or equal to the perpendicular distance from \(P\) to line \(BC\) (denoted as \(d(P, BC)\)).
   - This means that point \(P\) must lie inside or on the boundary of the cylinder \(\kappa_A\) with axis \(BC\) and passing through \(A\).

4. **Analogous conditions for other vertices:**
   - Similarly, point \(P\) must also lie inside or on the boundary of the cylinders \(\kappa_B\) and \(\kappa_C\) with axes \(CA\) and \(AB\) respectively, and passing through \(B\) and \(C\).

5. **Intersection of cylinders:**
   - Therefore, point \(P\) must lie inside or on the boundary of the intersection of the three cylinders \(\kappa_A\), \(\kappa_B\), and \(\kappa_C\).

6. **Projection onto the plane \(ABC\):**
   - The projection of the intersection of these cylinders onto the plane \(ABC\) forms a region. This region is known as the anticomplementary triangle \(\Delta\) of \(ABC\).

7. **Locus of point \(H\):**
   - It is important to note that no point \(H\) can lie on the boundary of \(\Delta\), but any interior point of \(\Delta\) can serve as \(H\) for a point \(P\) from \(\kappa_A \cap \kappa_B \cap \kappa_C\).
   - Thus, the locus of points \(H\) is the interior of the anticomplementary triangle \(\Delta\) without its boundary.

The final answer is the interior of the anticomplementary triangle \( \boxed{ \Delta } \) without its boundary.