Concept:Evaluate the determinant by using row operations and expansion, then equate to kabc to solve for k.Explanation:Start with the given determinant equation:a−a−a−bb−ba−b−c−a+b−c−a−b+c=kabcPerform row operation R2→R2+R1:a0−a−b0−ba−b−c−2c−a−b+c=kabcNow expand the determinant along the second row (using the row with two zeros). The only non‑zero term comes from the element in column 3: −2c.The minor sign for position (2,3) is (−1)2+3=−1.The corresponding cofactor is (−1)×a−a−b−b.Compute the 2×2 determinant:a−a−b−b=a(−b)−(−b)(−a)=−ab−ab=−2abThus the cofactor is (−1)×(−2ab)=2ab.Multiply by the entry −2c: (−2c)×(2ab)=−4abc.So the determinant simplifies to −4abc.Equating with the right‑hand side: −4abc=kabc.Since a, b, c are non‑zero, cancel abc to obtain k=−4.Answer:k=−4 (Option A)