Solve this

Solution:

Given: $\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|=0$

$\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|=0$

Applying $C_{1} \rightarrow C_{1}-C_{2}$

$\Rightarrow\left|\begin{array}{lll}x+1-x-2 & x+2 & x+a \\ x+2-x-3 & x+3 & x+b \\ x+3-x-4 & x+4 & x+c\end{array}\right|=0$

$\Rightarrow\left|\begin{array}{rrr}-1 & x+2 & x+a \\ -1 & x+3 & x+b \\ -1 & x+4 & x+c\end{array}\right|=0$

Taking out $(-1)$ common from $C_{1}$

$\Rightarrow(-1)\left|\begin{array}{lll}1 & x+2 & x+a \\ 1 & x+3 & x+b \\ 1 & x+4 & x+c\end{array}\right|=0$

Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$

$\Rightarrow(-1)\left|\begin{array}{ccc}1 & x+2 & x+a \\ 1-1 & x+3-x-2 & x+b-x-a \\ 1-1 & x+4-x-2 & x+c-x-a\end{array}\right|=0$

$\Rightarrow(-1)\left|\begin{array}{ccc}1 & x+2 & x+a \\ 0 & 1 & b-a \\ 0 & 2 & c-a\end{array}\right|=0$

Expanding through $C_{1}$

$\Rightarrow(-1)(1(c-a)-2(b-a))=0$

$\Rightarrow c-a-2 b+2 a=0$

$\Rightarrow c+a-2 b=0$

$\Rightarrow c+a=2 b$

$\Rightarrow a, b, c$ are in A.P.

Hence, if $\left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|=0$, then $a, b, c$ are in $\underline{\text { A.P. }}$.