The correct option is B 3
∣∣
∣
∣∣a3+1a2ba2cab2b3+1b2cac2bc2c3+1∣∣
∣
∣∣=11
⇒1abc∣∣
∣
∣∣a(a3+1)a3ba3cab3b(b3+1)b3cac3bc3c(c3+1)∣∣
∣
∣∣=11
⇒abcabc∣∣
∣
∣∣a3+1a3a3b3b3+1b3c3c3c3+1∣∣
∣
∣∣=11
Applying C1→C1−C3 and C2→C2−C3, we get
∣∣
∣
∣∣10a301b3−1−1c3+1∣∣
∣
∣∣=11
⇒a3+b3+c3+1=11
⇒a3+b3+c3=10
∴ Possible triplets are (1,1,2), (1,2,1) and (2,1,1)
Hence, number of triplets =3