Need Solution for R. D.  Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 54 Maths Textbook Solution.

Answer: $2$

Hint: Use determinant formula

Given:$\left|\begin{array}{ccc} a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c \end{array}\right|=0 \text { then using property of determinant. Find } \frac{a}{x}+\frac{b}{y}+\frac{c}{z} \text { where } x, y, z=0$

Solution:

\begin{aligned} &\text { Take } x, y, z \text { common from } C_{1}, C_{2}, C_{3}\\ &(x y z)\left|\begin{array}{ccc} \frac{a}{x} & \frac{b}{y}-1 & \frac{c}{z}-1 \\ a-\frac{1}{x} & \frac{b}{y} & \frac{c}{z}-1 \\ \frac{a}{x}-1 & \frac{b}{y}-1 & \frac{c}{z} \end{array}\right|=0 \end{aligned}

\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\mid \frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2 & \frac{b}{y}-1 & \frac{c}{z}-1 \\ &\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2 & \frac{b}{y} & \frac{c}{z}-1 \\ &\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2 & \frac{b}{y}-1 & \frac{c}{z} \end{aligned} \mid=0

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

\begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1} \\ &\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2\right)\left|\begin{array}{ccc} 1 & \frac{b}{y}-1 & \frac{c}{z}-1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|=0 \end{aligned}

$\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2\right) \cdot 1=0 \text { or } \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$