1. The problem states that if two rows or two columns of a matrix are equal, then the determinant of that matrix is zero. 2. Given the matrix: $$\begin{bmatrix} a & a & x \\ b & b & b \\ c & x & c \end{bmatrix}$$ 3. We want to find the value of $x$ such that the determinant of this matrix is zero. 4. Calculate the determinant using cofactor expansion along the first row: $$|A| = a \cdot \begin{vmatrix} b & b \\ x & c \end{vmatrix} - a \cdot \begin{vmatrix} b & b \\ c & c \end{vmatrix} + x \cdot \begin{vmatrix} b & b \\ c & x \end{vmatrix}$$ 5. Compute each minor: - $\begin{vmatrix} b & b \\ x & c \end{vmatrix} = b \cdot c - b \cdot x = b(c - x)$ - $\begin{vmatrix} b & b \\ c & c \end{vmatrix} = b \cdot c - b \cdot c = 0$ - $\begin{vmatrix} b & b \\ c & x \end{vmatrix} = b \cdot x - b \cdot c = b(x - c)$ 6. Substitute back: $$|A| = a \cdot b(c - x) - a \cdot 0 + x \cdot b(x - c) = a b (c - x) + x b (x - c)$$ 7. Factor $b$: $$|A| = b [a(c - x) + x(x - c)] = b [a c - a x + x^2 - x c]$$ 8. Group terms inside the bracket: $$a c - a x + x^2 - x c = x^2 - x(a + c) + a c$$ 9. So, $$|A| = b (x^2 - x(a + c) + a c)$$ 10. For the determinant to be zero, either $b=0$ or the quadratic equals zero: $$x^2 - x(a + c) + a c = 0$$ 11. Factor the quadratic: $$x^2 - x(a + c) + a c = (x - a)(x - c) = 0$$ 12. Therefore, $$x = a \quad \text{or} \quad x = c$$ 13. The value of $x$ must be either $a$ or $c$ for the determinant to be zero. 14. Hence, the correct answer is option d. $a$ or $c$.