The value of the determinant

$Δ=\begin{bmatrix}(a+1)(a+2) & a+2 & 1\\(a+2)(a+3) & a+3 & 1\\(a+3)(a+4) & a+4 & 1\end{bmatrix}$ is equal to

$-2$

The correct answer is option (1) : -2

Applying $R_2→R_2-R_1$ and $R_3→ R_r3-R_1,$ we obtain

$Δ=\begin{bmatrix}(a+1)(a+2) & a+2 & 1\\2(a+2) & 1 & 0\\2(a+5) & 2 & 0\end{bmatrix}$

$⇒4(a+2) -2(2a+5) = - 2 $         {On expanding along $C_3$]