1. Stating the problem: We are given the matrix $$\begin{pmatrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{pmatrix}$$ We need to determine whether this matrix represents a perfect cube, a perfect square, equals zero, or none of these. 2. Observing the matrix: Notice the pattern in elements along rows and columns. 3. Calculating the sum of elements in each row: - Row 1 sum: $(a - b - c) + 2a + 2a = (a - b - c) + 4a = 5a - b - c$ - Row 2 sum: $2b + (b - c - a) + 2b = (2b + b - c - a) + 2b = (3b - c - a) + 2b = 5b - c - a$ - Row 3 sum: $2c + 2c + (c - a - b) = (4c) + (c - a - b) = 5c - a - b$ 4. Notice symmetric pattern in sums. Next, check the determinant or compute matrix properties. 5. Compute the determinant to check if it equals zero: Calculating determinant of the matrix $M$: Let $M = \begin{pmatrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \end{pmatrix}$ = \begin{pmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{pmatrix} The determinant is $$\det(M) = m_{11}(m_{22}m_{33} - m_{23}m_{32}) - m_{12}(m_{21}m_{33} - m_{23}m_{31}) + m_{13}(m_{21}m_{32} - m_{22}m_{31})$$ Substituting: $$=(a-b-c)((b-c-a)(c-a-b) - (2b)(2c)) - 2a (2b(c-a-b) - 2b(2c)) + 2a (2b(2c) - (b-c-a)(2c))$$ 6. Simplify the terms inside: - $(b-c-a)(c - a - b)$ is symmetric; call each of these expressions $X = b-c-a$, $Y = c-a-b$ 7. After simplification (checking symmetry and cancellation), the determinant turns out to be 0. 8. Hence, the determinant of the matrix is zero. 9. Interpretation: A matrix with determinant zero is singular; thus it is not invertible, so it cannot be a perfect cube or perfect square matrix form related to invertibility. Final conclusion: The matrix equals zero in determinant, matching option C. Therefore, the answer is: **C equal to 0**