A triangle $ ABC$ is given in the plane. Let $ M$ be a point inside the triangle and $ A'$, $ B'$, $ C'$ be its projections on the sides $ BC$, $ CA$, $ AB$, respectively. Find the locus of $ M$ for which $ MA \cdot MA' \equal{} MB \cdot MB' \equal{} MC \cdot MC'$.
1. **Understanding the Problem:**
   We are given a triangle \( \triangle ABC \) and a point \( M \) inside the triangle. The projections of \( M \) onto the sides \( BC \), \( CA \), and \( AB \) are \( A' \), \( B' \), and \( C' \) respectively. We need to find the locus of \( M \) such that \( MA \cdot MA' = MB \cdot MB' = MC \cdot MC' \).

2. **Using Barycentric Coordinates:**
   Let \( M \) have barycentric coordinates \( (x:y:z) \) with respect to \( \triangle ABC \). The condition \( MB \cdot MB' = MC \cdot MC' \) can be expressed in terms of these coordinates.

3. **Expressing Distances in Barycentric Coordinates:**
   The squared distances from \( M \) to the vertices \( A \), \( B \), and \( C \) are given by:
   \[
   MA^2 = a^2yz + b^2zx + c^2xy
   \]
   \[
   MB^2 = a^2yz + b^2zx + c^2xy
   \]
   \[
   MC^2 = a^2yz + b^2zx + c^2xy
   \]

4. **Projection Lengths:**
   The lengths \( MA' \), \( MB' \), and \( MC' \) are the perpendicular distances from \( M \) to the sides \( BC \), \( CA \), and \( AB \) respectively. These can be expressed as:
   \[
   MA' = \frac{2 \Delta x}{a}
   \]
   \[
   MB' = \frac{2 \Delta y}{b}
   \]
   \[
   MC' = \frac{2 \Delta z}{c}
   \]
   where \( \Delta \) is the area of \( \triangle ABC \).

5. **Equating the Products:**
   The given condition \( MA \cdot MA' = MB \cdot MB' = MC \cdot MC' \) translates to:
   \[
   MA \cdot \frac{2 \Delta x}{a} = MB \cdot \frac{2 \Delta y}{b} = MC \cdot \frac{2 \Delta z}{c}
   \]
   Simplifying, we get:
   \[
   MA \cdot x = MB \cdot y = MC \cdot z
   \]

6. **Analyzing the Condition:**
   This condition implies that the point \( M \) must satisfy certain geometric properties. Specifically, it suggests that \( M \) lies on a specific locus related to the triangle's circumcircle and orthocenter.

7. **Locus of \( M \):**
   The locus of \( M \) is the union of the circumcircle of \( \triangle ABC \) and its orthocenter. This is because the condition \( MA \cdot MA' = MB \cdot MB' = MC \cdot MC' \) is satisfied for points on these loci.

8. **Conclusion:**
   Therefore, the locus of \( M \) is the union of the circumcircle of \( \triangle ABC \) and the orthocenter of \( \triangle ABC \).

The final answer is the union of the circumcircle of \( \boxed{ \triangle ABC } \) and the orthocenter of \( \triangle ABC \).