The correct option is B 4
△=∣∣
∣
∣∣b2+c2abacabc2+a2bccacba2+b2∣∣
∣
∣∣
Applying R1→aR1,R2→bR2,R3→cR3, we get △=1abc×∣∣
∣
∣∣a(b2+c2)a2ba2cab2b(c2+a2)cb2ac2bc2c(a2+b2)∣∣
∣
∣∣
Now, applying C1→1aC1,C2→1bC2,C3→1cC3, we get △=abcabc∣∣
∣
∣∣b2+c2a2a2b2c2+a2b2c2c2a2+b2∣∣
∣
∣∣
=∣∣
∣
∣∣0−2c2−2b2b2c2+a2b2c2c2a2+b2∣∣
∣
∣∣ [R1→R1−R2−R3]
Taking 2 common from R1 and applying R2→R2+R1, R3→R3+R1
=2∣∣
∣
∣∣0−c2−b2b2a20c20a2∣∣
∣
∣∣
Evaluating along R1, we get
△=2[c2(a2b2)−b2(−a2c2)]
=4a2b2c2
Hence, k=4.