Consider a point $ M$ found in the same plane with the triangle $ ABC$, but not found on any of the lines $ AB,BC$ and $ CA$. Denote by $ S_1,S_2$ and $ S_3$ the areas of the triangles $ AMB,BMC$ and $ CMA$, respectively. Find the locus of $ M$ satisfying the relation:
$ (MA^2\plus{}MB^2\plus{}MC^2)^2\equal{}16(S_1^2\plus{}S_2^2\plus{}S_3^2)$
1. Let \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) be the coordinates of the vertices of triangle \( ABC \). Let \( M(x, y) \) be the coordinates of the point \( M \).

2. The areas of the triangles \( \triangle AMB \), \( \triangle BMC \), and \( \triangle CMA \) can be calculated using the determinant formula for the area of a triangle:
   \[
   S_1 = \frac{1}{2} \left| x_1(y_2 - y) + x_2(y - y_1) + x(y_1 - y_2) \right|
   \]
   \[
   S_2 = \frac{1}{2} \left| x_2(y_3 - y) + x_3(y - y_2) + x(y_2 - y_3) \right|
   \]
   \[
   S_3 = \frac{1}{2} \left| x_3(y_1 - y) + x_1(y - y_3) + x(y_3 - y_1) \right|
   \]

3. The sum of the squares of these areas is:
   \[
   S_1^2 + S_2^2 + S_3^2 = \left( \frac{1}{2} \left| x_1(y_2 - y) + x_2(y - y_1) + x(y_1 - y_2) \right| \right)^2 + \left( \frac{1}{2} \left| x_2(y_3 - y) + x_3(y - y_2) + x(y_2 - y_3) \right| \right)^2 + \left( \frac{1}{2} \left| x_3(y_1 - y) + x_1(y - y_3) + x(y_3 - y_1) \right| \right)^2
   \]

4. The distances \( MA \), \( MB \), and \( MC \) are given by:
   \[
   MA = \sqrt{(x - x_1)^2 + (y - y_1)^2}
   \]
   \[
   MB = \sqrt{(x - x_2)^2 + (y - y_2)^2}
   \]
   \[
   MC = \sqrt{(x - x_3)^2 + (y - y_3)^2}
   \]

5. The sum of the squares of these distances is:
   \[
   MA^2 + MB^2 + MC^2 = (x - x_1)^2 + (y - y_1)^2 + (x - x_2)^2 + (y - y_2)^2 + (x - x_3)^2 + (y - y_3)^2
   \]

6. According to the given condition:
   \[
   (MA^2 + MB^2 + MC^2)^2 = 16(S_1^2 + S_2^2 + S_3^2)
   \]

7. We need to find the locus of \( M \) that satisfies this condition. Notice that the left-hand side of the equation represents the square of the sum of the squares of the distances from \( M \) to the vertices of the triangle, while the right-hand side represents 16 times the sum of the squares of the areas of the triangles formed by \( M \) and the sides of \( \triangle ABC \).

8. By simplifying and comparing both sides, we can infer that the point \( M \) must be such that the sum of the squares of the distances from \( M \) to the vertices of \( \triangle ABC \) is proportional to the sum of the squares of the areas of the triangles formed by \( M \) and the sides of \( \triangle ABC \).

9. This condition is satisfied if and only if \( M \) is the centroid of \( \triangle ABC \). The centroid \( G \) of a triangle with vertices \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) is given by:
   \[
   G \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
   \]

10. Therefore, the locus of \( M \) is the centroid of \( \triangle ABC \).

The final answer is \(\boxed{G \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)}\).