1. **State the problem:** We have a system of three equations: $$kx + 3y - z = 3$$ $$3x - y + z = -k$$ $$-16x - ky - kz = k$$ where $k$ is a positive constant. We need to find the value of $k$ such that the system has no unique solution. 2. **Recall the condition for no unique solution:** A system of linear equations has no unique solution if the determinant of the coefficient matrix is zero. 3. **Write the coefficient matrix:** $$A = \begin{bmatrix} k & 3 & -1 \\ 3 & -1 & 1 \\ -16 & -k & -k \end{bmatrix}$$ 4. **Calculate the determinant $\det(A)$:** $$\det(A) = k \begin{vmatrix} -1 & 1 \\ -k & -k \end{vmatrix} - 3 \begin{vmatrix} 3 & 1 \\ -16 & -k \end{vmatrix} - 1 \begin{vmatrix} 3 & -1 \\ -16 & -k \end{vmatrix}$$ 5. **Calculate each minor:** $$\begin{vmatrix} -1 & 1 \\ -k & -k \end{vmatrix} = (-1)(-k) - (1)(-k) = k + k = 2k$$ $$\begin{vmatrix} 3 & 1 \\ -16 & -k \end{vmatrix} = 3(-k) - 1(-16) = -3k + 16$$ $$\begin{vmatrix} 3 & -1 \\ -16 & -k \end{vmatrix} = 3(-k) - (-1)(-16) = -3k - 16$$ 6. **Substitute minors back:** $$\det(A) = k(2k) - 3(-3k + 16) - 1(-3k - 16) = 2k^2 + 9k - 48 + 3k + 16$$ 7. **Simplify:** $$\det(A) = 2k^2 + 12k - 32$$ 8. **Set determinant to zero for no unique solution:** $$2k^2 + 12k - 32 = 0$$ 9. **Divide entire equation by 2:** $$\cancel{2}k^2 + \cancel{2}6k - \cancel{2}16 = 0 \Rightarrow k^2 + 6k - 16 = 0$$ 10. **Solve quadratic equation:** $$k = \frac{-6 \pm \sqrt{6^2 - 4(1)(-16)}}{2} = \frac{-6 \pm \sqrt{36 + 64}}{2} = \frac{-6 \pm \sqrt{100}}{2}$$ $$k = \frac{-6 \pm 10}{2}$$ 11. **Calculate roots:** $$k_1 = \frac{-6 + 10}{2} = \frac{4}{2} = 2$$ $$k_2 = \frac{-6 - 10}{2} = \frac{-16}{2} = -8$$ 12. **Since $k$ is positive, the valid solution is:** $$\boxed{2}$$