In a plane we are given two distinct points $A,B$ and two lines $a, b$ passing through $B$ and $A$ respectively $(a  \ni  B, b  \ni A)$ such that the line $AB$ is equally inclined to a and b. Find the locus of points $M$ in the plane such that the product of distances from $M$ to $A$ and a equals the product of distances from $M$ to $B$ and $b$ (i.e., $MA \cdot  MA' = MB \cdot MB'$, where $A'$ and $B'$ are the feet of the perpendiculars from $M$ to $a$ and $b$ respectively).
1. **Given Conditions and Setup:**
   - Points \( A \) and \( B \) are distinct points in the plane.
   - Lines \( a \) and \( b \) pass through \( B \) and \( A \) respectively, such that \( a \ni B \) and \( b \ni A \).
   - The line \( AB \) is equally inclined to \( a \) and \( b \).
   - We need to find the locus of points \( M \) such that the product of distances from \( M \) to \( A \) and \( a \) equals the product of distances from \( M \) to \( B \) and \( b \).

2. **Generalization to Triangle:**
   - Consider a scalene triangle \( \triangle ABC \).
   - Let \( B' \) and \( C' \) be the orthogonal projections of \( M \) on the sidelines \( AC \) and \( AB \) respectively.
   - We need to find the locus of \( M \) such that \( \overline{MB} \cdot \overline{MB'} = \overline{MC} \cdot \overline{MC'} \).

3. **Barycentric Coordinates:**
   - Let \( M \) have barycentric coordinates \( (x:y:z) \) with respect to \( \triangle ABC \).
   - The distances from \( M \) to the sides of the triangle can be expressed in terms of these coordinates.

4. **Distance Product Condition:**
   - The condition \( \overline{MB} \cdot \overline{MB'} = \overline{MC} \cdot \overline{MC'} \) translates to:
     \[
     \frac{(a^2z^2 + 2S_Bzx + c^2x^2)y^2}{b^2} = \frac{(b^2x^2 + 2S_Cxy + a^2y^2)z^2}{c^2}
     \]
   - Simplifying this, we get:
     \[
     (a^2yz + b^2zx + c^2xy)(c^2xy + c^2yz - b^2yz - b^2xz) = 0
     \]

5. **Locus of Points \( M \):**
   - The above equation implies that \( M \) either lies on the circumcircle \( \mathcal{O} \) of \( \triangle ABC \) or on a rectangular circum-hyperbola \( \mathcal{H}_a \).
   - The circumcircle \( \mathcal{O} \) is given by:
     \[
     a^2yz + b^2zx + c^2xy = 0
     \]
   - The rectangular circum-hyperbola \( \mathcal{H}_a \) is the isogonal conjugate of the perpendicular bisector of \( \overline{BC} \).

The final answer is the union of the circumcircle of \( \boxed{ \triangle ABC } \) and a rectangular circum-hyperbola.