Let a, b, c, d be in arithmetic progression with common difference...

Since, $a, b, c \text{\text{ and } }$ d are in $A . P .$, therefore

$b - a = c - b = d - c = \lambda$

We have,

\left|$\begin{cases}x + a - c & x + b & x + a \\ x - 1 & x + c & x + b \\ x - b + d & x + d & x + c\end{cases}$\right| = 2$</p><p>${C}_{2} \to {C}_{2} - {C}_{3}$</p><p>$⇒ \left|$\begin{cases}x - 2\lambda & \lambda & x + a \\ x - 1 & \lambda & x + b \\ x + 2\lambda & \lambda & x + c\end{cases}$\right| = 2$</p><p>${R}_{2} \to {R}_{2} - {R}_{1}, {R}_{3} \to {R}_{3} - {R}_{1}</p><p>\left|\begin{cases}x - 2\lambda & \lambda & x + a \\ 2\lambda - 1 & 0 & b - a \\ 4\lambda & 0 & c - a\end{cases}\right| = 2

⇒ \lambda\left|\begin{cases}x - 2\lambda & 1 & x + a \\ 2\lambda - 1 & 0 & \lambda \\ 4\lambda & 0 & 2\lambda\end{cases}\right| = 2

$⇒ - \lambda\left(4{\lambda}^{2} - 4{\lambda}^{2} - 2\lambda\right) = 2$

$⇒ {\lambda}^{2} = 1$