(a) As given $\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 3}\\{x + 2}&{x + 3}&{x + 4}\\{x + a}&{x + b}&{x + c}\end{array}\,} \right|\, = \,0$

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