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Let's work through the questions one by one:

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### a) \[1p\] En \(\mathbb{R}^3\), el conjunto \( S = \{(x, y, z): xy \leq 0 \} \) es un subespacio vectorial de \(\mathbb{R}^3\).

**Analysis:**

A set is a vector subspace if it satisfies three properties:
1. Contains the zero vector.
2. Closed under vector addition.
3. Closed under scalar multiplication.

1. **Zero vector**: The zero vector \((0, 0, 0)\) satisfies the condition \( xy = 0 \), so it's in the set.

2. **Closed under addition**: Consider two vectors \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) such that \( x_1 y_1 \leq 0 \) and \( x_2 y_2 \leq 0 \). When adding, we need to check whether \( (x_1 + x_2)(y_1 + y_2) \leq 0 \). However, this isn't guaranteed because \( x_1 y_1 \leq 0 \) and \( x_2 y_2 \leq 0 \) doesn't imply that their sum will satisfy the same inequality.

3. **Closed under scalar multiplication**: Consider multiplying a vector \( (x, y, z) \) by a scalar \( c \). If \( xy \leq 0 \), then \( (cx)(cy) = c^2 xy \), which will always be greater than or equal to zero. If \( xy = 0 \), this condition holds. However, if \( xy < 0 \), multiplying by a scalar might violate the condition.

**Conclusion**: The set is not closed under addition or scalar multiplication, so it's not a subspace.

**Answer**: **False**.

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### b) \[2p\] El vector \( u = (3, 1, 2) \) pertenece al \( \text{Gen}\{(1, -1, 1), (2, 1, 1)\} \).

**Analysis**:

We need to check if the vector \( u = (3, 1, 2) \) is a linear combination of the vectors \( v_1 = (1, -1, 1) \) and \( v_2 = (2, 1, 1) \).

That is, we need to find scalars \( \alpha \) and \( \beta \) such that:
\[
\alpha (1, -1, 1) + \beta (2, 1, 1) = (3, 1, 2).
\]

This gives the system of equations:
1. \( \alpha + 2\beta = 3 \)
2. \( -\alpha + \beta = 1 \)
3. \( \alpha + \beta = 2 \)

Solving this system:

From equation (3):
\[
\alpha = 2 - \beta.
\]
Substitute into equation (2):
\[
-(2 - \beta) + \beta = 1 \implies -2 + 2\beta = 1 \implies \beta = \frac{3}{2}.
\]
Substitute \( \beta = \frac{3}{2} \) into \( \alpha = 2 - \beta \):
\[
\alpha = 2 - \frac{3}{2} = \frac{1}{2}.
\]

Now, substitute \( \alpha = \frac{1}{2} \) and \( \beta = \frac{3}{2} \) into equation (1) to verify:
\[
\frac{1}{2} + 2 \times \frac{3}{2} = \frac{1}{2} + 3 = 3.5 \neq 3.
\]

The system does not hold, so the vector \( u \) is **not** a linear combination of \( v_1 \) and \( v_2 \).

**Answer**: **False**.

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### c) \[1p\] El conjunto de vectores \( \{(2, 1, -1), (1, 1, 3), (-1, 2, 2)\} \) no es una base para \(\mathbb{R}^3\).

**Analysis**:

A set of vectors is a basis for \(\mathbb{R}^3\) if the vectors are linearly independent and span \(\mathbb{R}^3\). We can check for linear independence by forming a matrix with these vectors as rows or columns and calculating its determinant.

The matrix formed by these vectors is:
\[
\begin{pmatrix}
2 & 1 & -1 \\
1 & 1 & 3 \\
-1 & 2 & 2
\end{pmatrix}.
\]

We need to compute its determinant:
\[
\text{det} = 2 \begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 3 \\ -1 & 2 \end{vmatrix} + (-1) \begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix}.
\]
Calculating each minor:
\[
\begin{vmatrix} 1 & 3 \\ 2 & 2 \end{vmatrix} = (1)(2) - (3)(2) = 2 - 6 = -4,
\]
\[
\begin{vmatrix} 1 & 3 \\ -1 & 2 \end{vmatrix} = (1)(2) - (3)(-1) = 2 + 3 = 5,
\]
\[
\begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix} = (1)(2) - (1)(-1) = 2 + 1 = 3.
\]

Now substitute into the determinant formula:
\[
\text{det} = 2(-4) - 1(5) + (-1)(3) = -8 - 5 - 3 = -16.
\]

Since the determinant is non-zero (\(-16 \neq 0\)), the vectors are linearly independent and form a basis for \(\mathbb{R}^3\).

**Answer**: **False**.

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Do you have any questions or need more details?

Here are 5 related questions for further exploration:
1. What are the properties a set must satisfy to be a vector subspace?
2. How do we verify if a vector is a linear combination of other vectors?
3. How do we compute the determinant of a 3x3 matrix step-by-step?
4. Why does a non-zero determinant guarantee linear independence?
5. What is the geometric interpretation of a vector basis in \(\mathbb{R}^3\)?

**Tip**: When solving a system of linear equations, double-check your arithmetic to ensure the system is consistent or inconsistent.

Determine el valor de verdad (V o F) de cada una de las siguientes proposiciones. a) En ℝ^3, el conjunto S = {(x, y, z): xy ≤ 0} es un sub espacio vectorial de ℝ^3. b) El vector u = (3, 1, 2) pertenece al Gen{(1, -1, 1), (2, 1, 1)}. c) El conjunto de vectores {(2, 1, -1), (1, 1, 3), (-1, 2, 2)} no es una base para ℝ^3.

Analyze the properties of vector subspaces, check for linear combinations, and determine whether a set of vectors forms a basis for ℝ^3. This problem covers key concepts like subspace conditions, solving systems of equations, and using determinants to check for linear independence.

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