Answer:
Option C
Explanation:
Given a, b, c are in A.P.
.'. 2b = a+c
Now,$\begin{vmatrix}x+1& x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}$
[Applying R2 → 2R2]
= $\frac{1}{2}\begin{vmatrix}x+1& x+2 & x+a \\ 2x+4 & 2x+6 & 2x+2b \\ x+3 & x+4 & x+c \end{vmatrix}$
= $\frac{1}{2}\begin{vmatrix}x+1& x+2 & x+a \\ 2x+4 & 2x+6 & 2x+(a+c) \\ x+3 & x+4 & x+c \end{vmatrix}$
[Using equation(i)]
=$\frac{1}{2}\begin{vmatrix}x+1& x+2 & x+a \\ 0 & 0 & 0 \\ x+3 & x+4 & x+c \end{vmatrix}$
=$\frac{1}{2}.0$ =0
[Applying R2 → R2 - (R1+R3)]