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To solve the problem, we need to determine the values of \( k \) such that the vector \( \vec{u} = \begin{pmatrix} 2 \\ 1 \\ -5 \end{pmatrix} \) is in the space generated by the vectors \( \vec{v}_1 = \begin{pmatrix} 1 \\ 1 \\ k \end{pmatrix} \), \( \vec{v}_2 = \begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \), and \( \vec{v}_3 = \begin{pmatrix} 2 \\ k \\ 4 \end{pmatrix} \).

This means that \( \vec{u} \) can be written as a linear combination of \( \vec{v}_1 \), \( \vec{v}_2 \), and \( \vec{v}_3 \), i.e.,

\[
\vec{u} = \alpha \vec{v}_1 + \beta \vec{v}_2 + \gamma \vec{v}_3
\]

This equation can be written component-wise as the following system of equations:

\[
\begin{pmatrix} 2 \\ 1 \\ -5 \end{pmatrix} = \alpha \begin{pmatrix} 1 \\ 1 \\ k \end{pmatrix} + \beta \begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} + \gamma \begin{pmatrix} 2 \\ k \\ 4 \end{pmatrix}
\]

This leads to the following system of equations:

1. \( 2 = \alpha + \beta + 2\gamma \)
2. \( 1 = \alpha + 2\beta + k\gamma \)
3. \( -5 = \alpha k - 5\beta + 4\gamma \)

Now, we can solve this system of equations to find the values of \( k \). Let me calculate this for you.The solutions for \( \alpha \), \( \beta \), and \( \gamma \) in terms of \( k \) are:

\[
\alpha = \frac{5k + 22}{k^2 + k - 6}, \quad \beta = \frac{2k + 7}{k + 3}, \quad \gamma = \frac{-3k - 10}{k^2 + k - 6}
\]

For the system to have a valid solution, all the values of \( \alpha \), \( \beta \), and \( \gamma \) must be real numbers and finite, which means the denominators must not be zero. The denominator \( k^2 + k - 6 \) should not be zero, so let's solve for when this expression equals zero:

\[
k^2 + k - 6 = 0
\]

Let me calculate the roots of this equation.The equation \( k^2 + k - 6 = 0 \) has two solutions: \( k = -3 \) and \( k = 2 \).

These values of \( k \) would make the denominators in the solutions undefined, so they are not valid values for \( k \). Therefore, the possible values of \( k \) are all real numbers except \( k = -3 \) and \( k = 2 \).

Would you like further details on this, or any other questions?

Here are 5 related questions you could explore:

1. How do we interpret the meaning of the vector space spanned by the given vectors?
2. What happens geometrically when the determinant of the system's matrix is zero?
3. Can you solve a similar problem with different vectors?
4. How would this system change if \( \vec{u} \) were in higher dimensions?
5. How can you use row reduction to confirm the linear independence of the given vectors?

**Tip**: When solving systems of linear equations involving parameters, always check for values that make the denominator zero to avoid invalid solutions.

Determine los valores de k, si existen, de modo que el vector \( \vec{u} = \begin{pmatrix} 2 \\ 1 \\ -5 \end{pmatrix} \) esté en el espacio generado por los vectores \( \vec{v}_1 = \begin{pmatrix} 1 \\ 1 \\ k \end{pmatrix} \), \( \vec{v}_2 = \begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \), y \( \vec{v}_3 = \begin{pmatrix} 2 \\ k \\ 4 \end{pmatrix} \).

Determine the values of k that allow the vector \( \vec{u} = \begin{pmatrix} 2 \\ 1 \\ -5 \end{pmatrix} \) to be expressed as a linear combination of the vectors \( \vec{v}_1 = \begin{pmatrix} 1 \\ 1 \\ k \end{pmatrix} \), \( \vec{v}_2 = \begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \), and \( \vec{v}_3 = \begin{pmatrix} 2 \\ k \\ 4 \end{pmatrix} \). The solution involves solving a system of equations and a quadratic equation.

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