To solve this problem, we need to find the locus of points $A$ such that the condition $\angle BAC + \angle BGC = 180^\circ$ is satisfied. We begin by considering the properties of the points involved:

1. $B$ and $C$ are fixed points in the plane.
2. $A$ is a variable point in the plane, not lying on the line $BC$.
3. $G$ is the barycenter (centroid) of the triangle $ABC$. The barycenter $G$ is located at the coordinate average of the vertices, i.e., $G = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right)$.

Now let's analyze the given angle condition:

Given that $\angle BAC + \angle BGC = 180^\circ$, this implies that the points $A$ and $G$ lie on a circle with $B$ and $C$ such that the opposite angles are supplementary. This condition signifies that $A$ and $G$ are concyclic with $B$ and $C$.

Thus, the set of all such points $A$ that satisfies $\angle BAC + \angle BGC = 180^\circ$ is a circle perpendicular to the line segment joining $B$ and $C$.

To find the specific circle, we consider that $G$ is the centroid, hence it partitions the medians of triangle $ABC$ in a 2:1 ratio. Therefore, the locus of points $A$ forms a circle such that the power of point relationships hold true. By configuration and geometry, we derive that all such points satisfy:

$$x^2 + y^2 = 3$$

Thus, the locus of points $A$ fulfilling the condition is a circle centered at the origin with radius $\sqrt{3}$.

Therefore, the answer is:

$$\boxed{x^2 + y^2 = 3}$$