If a, b, c, are in A.P, then the determinant $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{x + 2}&{x + 3}&{x + 2a}\\{x + 3}&{x + 4}&{x + 2b}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right|$ is

(A) 0

(B) 1

(C) x

(D) 2x

Option a is correct

: $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{x + 2}&{x + 3}&{x + 2a}\\{x + 3}&{x + 4}&{x + 2b}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right|$

Applying ${R_1} \to {R_1} - {R_2},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&{ - 1}&{2a - 2b}\\{x + 3}&{x + 4}&{x + 2b}\\{x + 4}&{x + 5}&{x + 2c}\end{array}} \right|$

Applying ${C_1} \to {C_1} - {C_2},$

we get
$\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}0&{ - 1}&{2a - 2b}\\{ - 1}&{x + 4}&{x + 2b}\\{ - 1}&{x + 5}&{x + 2c}\end{array}} \right|$

Expanding along ${R_1},$

we get
$1\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&{x + 2b}\\{ - 1}&{x + 2c}\end{array}} \right| + (2a - 2b)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{ - 1}&{x + 4}\\{ - 1}&{x + 5}\end{array}} \right|$

$= - x - 2c + x + 2b + (2a - 2b)[ - x - 5 + x + 4]$
$= 2b - 2c + (2a - 2b)( - 1)$

$= 2b - 2c - 2a + 2b = 2[2b - (c + a)]$
$= 2\left[ {2\left( {\cfrac{{a + c}}{2}} \right) - (c + a)} \right] = 0$ [Since a, b, c are in A.P.]