If x is a positive integer, then

$\begin{vmatrix}x! & (x+1)! & (x+2)!\\(x+1)! & (x+2)! & (x+3)!\\(x+2)! & (x+3)! & (x+4)!\end{vmatrix}$ is equal to

$2x!(x+1)!(x+2)!$

The correct answer is option (2) : $2x!(x+1)!(x+2)!$

Let Δ be the value of the given determinant. Then,

$Δ=\begin{vmatrix}x! & (x+1)! & (x+2)!\\(x+1)x! & (x+2)(x+1)! & (x+3)(x+2)!\\(x+2)(x+1)x! & (x+3)(x+2)(x+1)! & (x+4)(x+3)(x+2)!\end{vmatrix}$

$⇒Δ=x!(x+1)!(x+2)!$

$\begin{vmatrix}1 & 1 & 1\\x+1 & x+2 & x+3\\(x+1)(x+2) & (x+2)(x+3) & (x+3)(x+4)\end{vmatrix}$

$⇒Δ=x!(x+1)!(x+2)!\begin{vmatrix}1 & 0 & 0\\x+1 & 1 & 2\\(x+1)(x+2) & 2(x+2) & 4x+10\end{vmatrix}$

$⇒Δ=2x!(x-1)!(x+2)!$