1. **State the problem:** We need to find all values of $k$ such that the product $$[k \quad 1 \quad 1] \times \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & -3 \end{bmatrix} \times \begin{bmatrix} k \\ 1 \\ 1 \end{bmatrix} = 0$$ 2. **Rewrite the expression:** Let $\mathbf{v} = \begin{bmatrix} k \\ 1 \\ 1 \end{bmatrix}$ and $\mathbf{w} = [k \quad 1 \quad 1]$. The expression is $\mathbf{w} \times A \times \mathbf{v} = 0$ where $$A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & -3 \end{bmatrix}$$ 3. **Calculate $A \mathbf{v}$:** $$A \mathbf{v} = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 2 \\ 0 & 2 & -3 \end{bmatrix} \begin{bmatrix} k \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot k + 1 \cdot 1 + 0 \cdot 1 \\ 1 \cdot k + 0 \cdot 1 + 2 \cdot 1 \\ 0 \cdot k + 2 \cdot 1 + (-3) \cdot 1 \end{bmatrix} = \begin{bmatrix} k + 1 \\ k + 2 \\ 2 - 3 \end{bmatrix} = \begin{bmatrix} k + 1 \\ k + 2 \\ -1 \end{bmatrix}$$ 4. **Calculate $\mathbf{w} (A \mathbf{v})$:** $$[k \quad 1 \quad 1] \times \begin{bmatrix} k + 1 \\ k + 2 \\ -1 \end{bmatrix} = k(k + 1) + 1(k + 2) + 1(-1)$$ 5. **Simplify the expression:** $$k(k + 1) + (k + 2) - 1 = k^2 + k + k + 2 - 1 = k^2 + 2k + 1$$ 6. **Set the expression equal to zero and solve for $k$:** $$k^2 + 2k + 1 = 0$$ This is a quadratic equation which factors as: $$ (k + 1)^2 = 0 $$ 7. **Find the solution:** $$k = -1$$ **Final answer:** The only value of $k$ that satisfies the equation is $\boxed{-1}$.