$D=\begin{vmatrix}x+1&x+2&x+a\\x+2&x+3&x+b\\x+3&x+4&x+c\end{vmatrix}=0$ if $a,b,c$ are in
Step-by-Step Solution
Key Concept: Apply $R_2-R_1$ and $R_3-R_2$; simplify
$R_2-R_1:(1,1,b-a)$, $R_3-R_1:(2,2,c-a)$. $D=\begin{vmatrix}x+1&x+2&x+a\\1&1&b-a\\2&2&c-a\end{vmatrix}$. $C_2-C_1$: $(1,0,0)$ col 2. Expand: determinant vanishes iff $2(b-a)-(c-a)=0\Rightarrow 2b-2a=c-a\Rightarrow c+a=2b$ (AP).
Correct Answer: 4