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Need Solution for R. D.  Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 52 sub question 1 Maths Textbook Solution.

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Answer: 0

Hint Use determinant formula

Given: $$ \left|\begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{array}\right|=0

Solution:

$$ \begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\left|\begin{array}{lcc} x+a+b+c & b & c \\ x+a+b+c & x+b & c \\ x+a+b+c & b & x+c \end{array}\right| \\ &(x+a+b+c) \text { common from } \mathrm{C}_{1} \\ &=(x+a+b+c)\left|\begin{array}{ccc} 1 & b & c \\ 1 & x+b & c \\ 1 & b & x+c \end{array}\right| \end{aligned}

$$ \begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1}\\ &=(x+a+b+c)\left|\begin{array}{lll} 1 & b & c \\ 0 & x & 0 \\ 0 & 0 & x \end{array}\right|\\ &\text { Expand from } \mathrm{C}_{1}\\ &=(x+a+b+c)\left(\left(x^{2}-0\right)-0+0\right)=0\\ &x+a+b+c=0, x^{2}=0\\ &x=-(a+b+c), x=0 \end{aligned}

 

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