The value of $$\begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix}$$ is

We evaluate the determinant $$\begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix}$$ using row operations to simplify.

Apply $$R_2 \to R_2 - R_1$$: the first entry becomes $$(a+2)(a+3) - (a+1)(a+2) = (a+2)[(a+3) - (a+1)] = 2(a+2)$$, the second entry becomes $$(a+3) - (a+2) = 1$$, and the third entry becomes $$1 - 1 = 0$$.

Apply $$R_3 \to R_3 - R_2$$ (using the original $$R_2$$ and $$R_3$$): the first entry becomes $$(a+3)(a+4) - (a+2)(a+3) = (a+3)[(a+4) - (a+2)] = 2(a+3)$$, the second entry becomes $$(a+4) - (a+3) = 1$$, and the third entry becomes $$1 - 1 = 0$$.

The transformed determinant is $$\begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ 2(a+2) & 1 & 0 \\ 2(a+3) & 1 & 0 \end{vmatrix}$$.

Expanding along the third column (since it has two zeros), the determinant equals $$1 \cdot \begin{vmatrix} 2(a+2) & 1 \\ 2(a+3) & 1 \end{vmatrix}$$, with the sign factor $$(-1)^{1+3} = +1$$ for the $$(1,3)$$ cofactor.

This $$2 \times 2$$ determinant evaluates to $$2(a+2)(1) - 1 \cdot 2(a+3) = 2(a+2) - 2(a+3) = 2a + 4 - 2a - 6 = -2$$.

Therefore the value of the determinant is $$-2$$.