The correct option is B a2(a+z+x+y)
We have, ∣∣
∣∣a+xyzxa+yzxya+z∣∣
∣∣
Applying C1→C1+C2+C3=∣∣
∣∣a+x+y+zyza+x+y+za+yza+x+y+zya+z∣∣
∣∣=(a+x+y+z)∣∣
∣∣1yz1a+yz1ya+z∣∣
∣∣
Applying R2→R2−R1 and R3→R3−R1=(a+x+y+z)∣∣
∣∣1yz0a000a∣∣
∣∣=(a+x+y+z)∣∣∣a00a∣∣∣=a2(a+z+x+y)