Answer:
Option C
Explanation:
Given a, b, c are in A.P.
.'. 2b = a+c
Now,|x+1x+2x+ax+2x+3x+bx+3x+4x+c|
[Applying R2 → 2R2]
= 12|x+1x+2x+a2x+42x+62x+2bx+3x+4x+c|
= 12|x+1x+2x+a2x+42x+62x+(a+c)x+3x+4x+c|
[Using equation(i)]
=12|x+1x+2x+a000x+3x+4x+c|
=12.0 =0
[Applying R2 → R2 - (R1+R3)]