For what value of $k$, the following system have a unique solution? (where $\mathbb{R}$ is set of real numbers)

$x + y + z = 1$

$2x + 3y + 4z = 3$

$x - y + kz = 5$

A system of linear equations has a unique solution when the determinant of the coefficient matrix is not equal to zero.

The coefficient matrix from the three equations is:

$A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 1 & -1 & k \end{bmatrix}$

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Using first row expansion to calculate the determinant:

$\det(A) = 1 \times \begin{vmatrix} 3 & 4 \\ -1 & k \end{vmatrix} - 1 \times \begin{vmatrix} 2 & 4 \\ 1 & k \end{vmatrix} + 1 \times \begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix}$

Computing each 2×2 determinant:

$\begin{vmatrix} 3 & 4 \\ -1 & k \end{vmatrix} = 3k + 4$

$\begin{vmatrix} 2 & 4 \\ 1 & k \end{vmatrix} = 2k - 4$

$\begin{vmatrix} 2 & 3 \\ 1 & -1 \end{vmatrix} = -5$

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Substituting back:

$\det(A) = 1(3k + 4) - 1(2k - 4) + 1(-5)$

$= 3k + 4 - 2k + 4 - 5$

$= k + 3$

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For a unique solution, the determinant must be non-zero:

$k + 3 \neq 0$

$k \neq -3$

Therefore, the system has a unique solution for all real numbers except $k = -3$.

The answer is $k \in \mathbb{R} \setminus \{-3\}$