If x, y, z are in A.P., then the value of the determinant

$\begin{vmatrix} a+ 2& a+3& a+ 2x\\a+ 3& a+ 4& a+ 2y \\a+ 4 & a + 5 & a= 2z \end{vmatrix},$ is

0

The correct answer is option (1) : 0

It is given that $x, y, z $ are in A.P.

$∴x+z-2y = 0 $

Now,

$\begin{vmatrix} a+ 2& a+3& a+ 2x\\a+ 3& a+ 4& a+ 2y \\a+ 4 & a + 5 & a= 2z \end{vmatrix}$

$\begin{vmatrix} 0& 0 & 2(x+z-2y)\\a+ 3& a+ 4& a+ 2y \\a+ 4 & a + 5 & a= 2z \end{vmatrix}$ [Applying $R_1→R_1+R_3-2R_2$]

$=\begin{vmatrix} 0& 0 & 0\\a+ 3& a+ 4& a+ 2y \\a+ 4 & a + 5 & a= 2z \end{vmatrix}=0$          $[∵x+z-2y =0]$