If $p, q, r$ and $s$ are in AP and 

$f\left(x\right)=\left|\begin{array}{ccc}p+\sin x& q+\sin x& p-r+\sin x\\ q+\sin x& r+\sin x& -1+\sin x\\ r+\sin x& s+\sin x& s-q+\sin x\end{array}\right|$ such that  ${\int }_0^1 f\left(x\right)dx=-2$ , the common difference of the AP can be

$\because f\left(x\right)=\frac{1}{2}\left|\begin{array}{ccc}p+\sin x& q+\sin x& p-r+\sin x\\ 2q+2\sin x& 2r+2\sin x& -2+2\sin x\\ r+\sin x& s+\sin x& s-q+\sin x\end{array}\right|$

Applying $R_2\to R_2-\left(R_1+R_3\right)$, then 

$f\left(x\right)=\frac{1}{2}\left|\begin{array}{ccccc}p+\sin x& & q+\sin x& p-r+\sin x& \\ & & & & ⋮\\ 0& \cdots & 0& \cdots & -2\\ r+\sin x& & s+\sin x& & s-q+\sin x\end{array}\right|$

$\begin{array}{l}[\because 2q=p+r,2r=q+s\text{ and }p+s=q+r]\\ =-\frac{(-2)}{2}\left|\begin{array}{ll}p+\sin x& q+\sin x\\ r+\sin x& s+\sin x\end{array}\right|\end{array}$

Applying $C_2\to C_2-C_1$, then

$f\left(x\right)=\left|\begin{array}{cc}p+\sin x& D\\ p+2D+\sin x& D\end{array}\right|$

                                [where D = common difference]

            $=D[p+\sin x-p-2D-\sin x]=-2D^2$

and                ${\int }_0^1 f\left(x\right)dx=-4$

$\begin{array}{l}\Rightarrow {\int }_0^1 \left(-2D^2\right)dx=-4\Rightarrow -2D^2=-2\\ \therefore D^2=1\Rightarrow D=\pm 1\end{array}$