Determine and construct the set of points $M$ in the plane $P$ such that $||-2\vec{MA} + \vec{MB} + 3\vec{MC}|| = 2\sqrt{10}$.

1. Problem Description

Determine and construct the set of points $M$ in the plane $P$ such that $||-2\vec{MA} + \vec{MB} + 3\vec{MC}|| = 2\sqrt{10}$.

2. Solution Steps

Let $G$ be the barycenter of the points $A, B, C$ with weights $-2, 1, 3$ respectively. Then, by definition of the barycenter:

$-2\vec{GA} + \vec{GB} + 3\vec{GC} = \vec{0}$

Now, we can rewrite the expression inside the norm:

$-2\vec{MA} + \vec{MB} + 3\vec{MC} = -2(\vec{MG} + \vec{GA}) + (\vec{MG} + \vec{GB}) + 3(\vec{MG} + \vec{GC}) = -2\vec{MG} - 2\vec{GA} + \vec{MG} + \vec{GB} + 3\vec{MG} + 3\vec{GC} = (-2+1+3)\vec{MG} + (-2\vec{GA} + \vec{GB} + 3\vec{GC}) = 2\vec{MG} + \vec{0} = 2\vec{MG}$

So, the given equation becomes:

$||2\vec{MG}|| = 2\sqrt{10}$

$2||\vec{MG}|| = 2\sqrt{10}$

$||\vec{MG}|| = \sqrt{10}$

$MG = \sqrt{10}$

The set of points $M$ such that $MG = \sqrt{10}$ is a circle with center $G$ and radius $\sqrt{10}$.

To find the coordinates of $G$, we can use the formula for the barycenter:

$\vec{OG} = \frac{-2\vec{OA} + \vec{OB} + 3\vec{OC}}{-2+1+3} = \frac{-2\vec{OA} + \vec{OB} + 3\vec{OC}}{2}$

So $G$ is the barycenter of $A, B, C$ with weights $-2, 1, 3$ respectively. The set of points $M$ is the circle centered at $G$ with radius $\sqrt{10}$.

3. Final Answer

The set of points $M$ is a circle with center $G$, the barycenter of $A, B, C$ with weights $-2, 1, 3$, and radius $\sqrt{10}$.