The value of the determinant

$Δ=\begin{vmatrix}x+2 & x+3 & x+ 5\\x+4 & x+ 6 & x+ 9 \\x+ 8 & x+ 11 & x + 15 \end{vmatrix}$, is

-2

The correct answer is option (2) : -2

Applying $R_2→R_2-R_1$ and $R_3→R_3-R_2,$ we get

$Δ=\begin{vmatrix}x+2 & x+3 & x+ 5\\2  &3 & 4 \\4 & 5 & 6 \end{vmatrix}$

$⇒Δ=2\begin{vmatrix}x & x & x+ 1\\2  &3 & 4 \\1 & 1 & 1 \end{vmatrix}$    $\begin{bmatrix}Applying \, R_1→R_1-R_2\\and\, R_3 → R_3-R_2\end{bmatrix}$

$⇒Δ=2\begin{vmatrix}x & 0 & 1\\2  &1 & 1 \\1 & 0 & 0 \end{vmatrix}$    $\begin{bmatrix}Applying \, C_2→C_2-C_1\\and\, C_3 → C_3-C_2\end{bmatrix}$

$⇒Δ=-2$