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

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

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

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

$⇒Δ=x!(x + 1)! (x + 2)!$

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

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

$⇒Δ=2x!(x+1)! (x + 2)!$