Let S be the set of all real values of k for which the system of linear equations
$$x + y + z = 2$$
$$2x + y - z = 3$$
$$3x + 2y + kz = 4$$
has a unique solution. Then S is:

We are asked to find all real numbers $$k$$ for which the given system of three linear equations

$$\begin{aligned} x + y + z &= 2 \\ 2x + y - z &= 3 \\ 3x + 2y + kz &= 4 \end{aligned}$$

has a unique solution. A system of three linear equations in three unknowns has a unique solution exactly when the determinant of its coefficient matrix is non-zero. We first form that coefficient matrix:

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

The determinant of a $$3\times3$$ matrix $$ \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix} $$ is given by the formula

$$ \det A \;=\; a_{11}(a_{22}a_{33}-a_{23}a_{32}) \;-\; a_{12}(a_{21}a_{33}-a_{23}a_{31}) \;+\; a_{13}(a_{21}a_{32}-a_{22}a_{31}). $$

We now apply this formula to our matrix $$A$$, identifying each entry:

$$ \begin{aligned} a_{11}&=1, & a_{12}&=1, & a_{13}&=1,\\ a_{21}&=2, & a_{22}&=1, & a_{23}&=-1,\\ a_{31}&=3, & a_{32}&=2, & a_{33}&=k. \end{aligned} $$

Substituting these values into the determinant formula, we obtain

$$ \det A =\;1\bigl(1\cdot k-(-1)\cdot2\bigr) -\;1\bigl(2\cdot k-(-1)\cdot3\bigr) +\;1\bigl(2\cdot2-1\cdot3\bigr). $$

We now simplify each of the three products step by step.

First product (inside the first parentheses):

$$1\cdot k-(-1)\cdot2 =\;k+2.$$

Second product (inside the second parentheses):

$$2\cdot k-(-1)\cdot3 =\;2k+3.$$

Third product (inside the third parentheses):

$$2\cdot2-1\cdot3 =\;4-3 =\;1.$$

Substituting these simplified results back into the expression for the determinant, we have

$$ \det A =\;1(k+2)\;-\;1(2k+3)\;+\;1(1). $$

Because each coefficient in front of the parentheses is $$1$$ (with the appropriate sign), we can drop the explicit multiplication:

$$ \det A =\;(k+2)\;-\;(2k+3)\;+\;1. $$

Now we simplify the right-hand side term by term:

$$ \begin{aligned} (k+2)&-(2k+3)+1 &=&\;k+2-2k-3+1\\ &=&\;(k-2k)+(2-3+1)\\ &=&\;-k+0\\ &=&\;-k. \end{aligned} $$

Thus, the determinant of matrix $$A$$ is

$$ \det A = -k. $$

For the system to possess a unique solution, we require that this determinant be non-zero:

$$ \det A \neq 0 \quad\Longrightarrow\quad -k \neq 0 \quad\Longrightarrow\quad k \neq 0. $$

All real numbers except zero satisfy this condition. Therefore, the set $$S$$ of all real values of $$k$$ for which the system has a unique solution is

$$ S = \mathbb{R}\setminus\{0\}. $$

Hence, the correct answer is Option B.