1. **State the problem:** We have the system of linear equations: $$x - ky + z = 0,$$ $$kx + 3y - kz = 0,$$ $$3x + y - z = 0.$$ Find all values of $k$ such that the trivial solution $(x,y,z) = (0,0,0)$ is the only solution. 2. **Write the augmented matrix and coefficient matrix:** The coefficient matrix is $$A=\begin{bmatrix} 1 & -k & 1 \\ k & 3 & -k \\ 3 & 1 & -1 \end{bmatrix}.$$ 3. **Condition for trivial solution only:** The system has only the trivial solution iff $\det(A) \neq 0$. So we find $\det(A)$: $$\det(A) = 1 \times \begin{vmatrix} 3 & -k \\ 1 & -1 \end{vmatrix} - (-k) \times \begin{vmatrix} k & -k \\ 3 & -1 \end{vmatrix} + 1 \times \begin{vmatrix} k & 3 \\ 3 & 1 \end{vmatrix}.$$ 4. **Calculate the minors:** - First minor: $$3 \times (-1) - (-k) \times 1 = -3 + k = k - 3.$$ - Second minor: $$k \times (-1) - (-k) \times 3 = -k + 3k = 2k.$$ - Third minor: $$k \times 1 - 3 \times 3 = k - 9.$$ 5. **Compute the determinant:** $$\det(A) = 1 \times (k - 3) + k \times 2k + 1 \times (k - 9) = (k - 3) + 2k^2 + (k - 9).$$ 6. **Simplify:** $$\det(A) = 2k^2 + (k + k) - 3 - 9 = 2k^2 + 2k - 12.$$ 7. **Set determinant not equal to zero:** $$2k^2 + 2k - 12 \neq 0,$$ or divide both sides by 2: $$k^2 + k - 6 \neq 0.$$ 8. **Factor the quadratic:** $$k^2 + k - 6 = (k + 3)(k - 2).$$ 9. **Find values for which determinant equals zero:** $$ (k+3)(k-2) = 0 \to k = -3 \text{ or } k=2.$$ 10. **Conclusion:** The trivial solution is unique iff $k \neq 2$ and $k \neq -3$. Thus the set of all values of $k$ is $\mathbb{R} - \{2, -3\}$. **Final answer:** B $\mathbb{R} - \{2, -3\}$.