If a, b, c are in A.P., then the value determinant $ $\begin{vmatrix} x+2 & x+3 & x+2a\\x+3 & x+4 &x+2b \\x+4 & x+5 & x+2c\end{vmatrix}$ is :

0

The correct answer is Option (1) → 0

a, b, c are in A.P.

$⇒c-b=b-a$= say(d)

or $a+c=2b$

$Δ=\begin{vmatrix} x+2 & x+3 & x+2a\\x+3 & x+4 &x+2b \\x+4 & x+5 & x+2c\end{vmatrix}$

$⇒R_3→R_3-R_2$

$\begin{vmatrix} x+2 & x+3 & x+2a\\x+3 & x+4 &x+2b \\1 & 1 & 2(c-b)\end{vmatrix}$

$⇒R_2→R_2-R_1$

$\begin{vmatrix} x+2 & x+3 & x+2a\\1 & 1 &2(b-a) \\1 & 1 & 2(c-b)\end{vmatrix}$

$2(b-a)=2(c-b)$

so $R_2=R_3⇒Δ=0$