= $\left| {\,\begin{array}{*{20}{c}}{ - 1}&{ - 1}&{x + 3}\\{ - 1}&{ - 1}&{x + 4}\\{a - b}&{b - c}&{x + c}\end{array}\,} \right|\, = 0$, by $\begin{array}{l}{C_1} \to {C_1} - {C_2}\\{C_2} \to {C_2} - {C_3}\end{array}$

$ \Rightarrow $ $\left| {\,\begin{array}{*{20}{c}}0&0&{ - 1}\\{ - 1}&{ - 1}&{x + 4}\\{a - b}&{b - c}&{x + c}\end{array}\,} \right|\, = 0$, by ${R_1} \to {R_1} - {R_2}$

$ \Rightarrow $$( - 1)\,( - b + c + a - b)\, = 0$

$ \Rightarrow $ $2b - a - c = 0 \Rightarrow a + c = 2b$ i.e., $a,b,c$ are in $A.P.$

Trick : In such type of problem, put any suitable value of $x$ i.e. $0$, so that the determinant.

$\left| {\,\begin{array}{*{20}{c}}1&2&3\\2&3&4\\a&b&c\end{array}\,} \right| = 0$

$ \Rightarrow 1\,(3c - 4b) - 2(2c - 4a) + 3(2b - 3a) = 0$

$ \Rightarrow $ $ - c + 2b - a = 0 \Rightarrow 2b = a + c$. Hence the result.