#### Explain Solution R.D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 48 maths Textbook Solution.

Answer: $0$

Hint $\alpha, \beta, \gamma \text { are in AP }$

Given:$\left|\begin{array}{lll} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right|=0 \text { where } \alpha, \beta, \gamma \text { are in } \mathrm{AP}$

Solution:

\begin{aligned} &\alpha, \beta, \gamma \text { are in } \mathrm{AP}\\ &a_{2}-a_{1}=a_{3}-a_{1}\\ &\beta-\gamma=\gamma-\beta\\ &\beta+\beta=\gamma+\alpha\\ &2 \beta=\gamma+\alpha\\ &\gamma+\alpha-2 \beta=0 \end{aligned}                                ...(1)

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \\ &\text { Now } R_{1} \rightarrow R_{1}+R_{3}-2 R_{2} \\ &=\left|\begin{array}{ccc} 0 & 0 & 2 \beta-(\alpha+\gamma) \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \end{aligned}

\begin{aligned} &=\left|\begin{array}{ccc} 0 & 0 & 2 \beta-2 \beta \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \ldots \ldots \ldots \text { from(1) } \\ &=\left|\begin{array}{ccc} 0 & 0 & 0 \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \end{aligned}

Expanding along $R_{1}$,

$=0$ ..hence proved