We know that $R^3 = \{(x_1, x_2, x_3) | x_i \in R, i = 1, 2, 3\}$ is a vector space regarding the laws of composition
$(x_1, x_2, x_3) + (y_1, y_2, y_3) = (x_1 + y_1, x_2 + y_2, x_3 + y_3)$, $\lambda (x_1, x_2, x_3) = (\lambda x_1, \lambda x_2, \lambda x_3)$, $\lambda \in R$.
We consider the following subset of $R^3$ : $L =\{(x_1, x2, x_3) \in R^3 | x_1 + x_2 + x_3 = 0\}$.
a) Prove that $L$ is a vector subspace of $R^3$ .
b) In $R^3$ the following relation is defined $\overline{x} R \overline{y}  \Leftrightarrow \overline{x} -\overline{y} \in L, \overline{x} , \overline{y} \in R^3$.
Prove that it is an equivalence relation.
c) Find two vectors of $R^3$ that belong to the same class as the vector $(-1, 3, 2)$.
### Part (a)
To prove that \( L \) is a vector subspace of \( \mathbb{R}^3 \), we need to show that \( L \) satisfies the following properties:
1. The zero vector is in \( L \).
2. \( L \) is closed under vector addition.
3. \( L \) is closed under scalar multiplication.

1. **Zero Vector**: 
   The zero vector in \( \mathbb{R}^3 \) is \( (0, 0, 0) \). We check if it belongs to \( L \):
   \[
   0 + 0 + 0 = 0
   \]
   Hence, \( (0, 0, 0) \in L \).

2. **Closed under Vector Addition**: 
   Let \( \mathbf{u} = (u_1, u_2, u_3) \in L \) and \( \mathbf{v} = (v_1, v_2, v_3) \in L \). Then:
   \[
   u_1 + u_2 + u_3 = 0 \quad \text{and} \quad v_1 + v_2 + v_3 = 0
   \]
   We need to show that \( \mathbf{u} + \mathbf{v} \in L \):
   \[
   (u_1 + v_1) + (u_2 + v_2) + (u_3 + v_3) = (u_1 + u_2 + u_3) + (v_1 + v_2 + v_3) = 0 + 0 = 0
   \]
   Therefore, \( \mathbf{u} + \mathbf{v} \in L \).

3. **Closed under Scalar Multiplication**: 
   Let \( \mathbf{u} = (u_1, u_2, u_3) \in L \) and \( \lambda \in \mathbb{R} \). Then:
   \[
   u_1 + u_2 + u_3 = 0
   \]
   We need to show that \( \lambda \mathbf{u} \in L \):
   \[
   \lambda (u_1, u_2, u_3) = (\lambda u_1, \lambda u_2, \lambda u_3)
   \]
   \[
   \lambda u_1 + \lambda u_2 + \lambda u_3 = \lambda (u_1 + u_2 + u_3) = \lambda \cdot 0 = 0
   \]
   Therefore, \( \lambda \mathbf{u} \in L \).

Since \( L \) satisfies all three properties, \( L \) is a vector subspace of \( \mathbb{R}^3 \).

### Part (b)
To prove that the relation \( \overline{x} R \overline{y} \Leftrightarrow \overline{x} - \overline{y} \in L \) is an equivalence relation, we need to show that it is reflexive, symmetric, and transitive.

1. **Reflexive**: 
   For any \( \overline{x} \in \mathbb{R}^3 \):
   \[
   \overline{x} - \overline{x} = (x_1, x_2, x_3) - (x_1, x_2, x_3) = (0, 0, 0) \in L
   \]
   Hence, \( \overline{x} R \overline{x} \).

2. **Symmetric**: 
   If \( \overline{x} R \overline{y} \), then \( \overline{x} - \overline{y} \in L \). This implies:
   \[
   \overline{x} - \overline{y} = (x_1 - y_1, x_2 - y_2, x_3 - y_3) \in L
   \]
   Since \( L \) is a subspace, it is closed under negation:
   \[
   \overline{y} - \overline{x} = -( \overline{x} - \overline{y} ) \in L
   \]
   Hence, \( \overline{y} R \overline{x} \).

3. **Transitive**: 
   If \( \overline{x} R \overline{y} \) and \( \overline{y} R \overline{z} \), then \( \overline{x} - \overline{y} \in L \) and \( \overline{y} - \overline{z} \in L \). We need to show that \( \overline{x} R \overline{z} \):
   \[
   \overline{x} - \overline{z} = (\overline{x} - \overline{y}) + (\overline{y} - \overline{z})
   \]
   Since \( L \) is closed under addition:
   \[
   (\overline{x} - \overline{y}) + (\overline{y} - \overline{z}) \in L
   \]
   Hence, \( \overline{x} R \overline{z} \).

Since the relation is reflexive, symmetric, and transitive, it is an equivalence relation.

### Part (c)
To find two vectors in \( \mathbb{R}^3 \) that belong to the same equivalence class as \( (-1, 3, 2) \), we need to find vectors \( \overline{v} \) such that \( (-1, 3, 2) - \overline{v} \in L \). This means:
\[
(-1, 3, 2) - (v_1, v_2, v_3) = (w_1, w_2, w_3) \quad \text{where} \quad w_1 + w_2 + w_3 = 0
\]
Let's choose \( \overline{v} = (0, 0, 4) \):
\[
(-1, 3, 2) - (0, 0, 4) = (-1, 3, -2)
\]
\[
-1 + 3 - 2 = 0
\]
So, \( (0, 0, 4) \) is in the same equivalence class as \( (-1, 3, 2) \).

Another vector, let's choose \( \overline{v} = (1, 2, 0) \):
\[
(-1, 3, 2) - (1, 2, 0) = (-2, 1, 2)
\]
\[
-2 + 1 + 1 = 0
\]
So, \( (1, 2, 0) \) is in the same equivalence class as \( (-1, 3, 2) \).

The final answer is \( \boxed{ (0, 0, 4) } \) and \((1, 2, 0)\)