#### Explain Solution R. D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 50 maths Textbook Solution.

Answer: $2$

Hint Use determinant formula

Given: $\left|\begin{array}{lll} p & b & c \\ q & q & c \\ a & b & r \end{array}\right|=0 \text { Find value of } \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} \& p \neq a, q \neq b, r \neq c$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} p & b & c \\ q & q & c \\ a & b & r \end{array}\right| \\ &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}-R_{3} \& R_{2} \rightarrow R_{2}-R_{3} \\ &\left|\begin{array}{ccc} p-a & 0 & c-p \\ 0 & q-b & c-p \\ a & 0 & r \end{array}\right|=0 \end{aligned}

\begin{aligned} &\text { Expand from } \mathrm{R}_{1}\\ &=p-a\left|\begin{array}{cc} q-b & c-p \\ b & r \end{array}\right|+(c-r)\left|\begin{array}{cc} 0 & q-b \\ a & b \end{array}\right|=0\\ &=(p-a)[r(q-b)-b(c-r)+(c-r)(-a(q-b))]=0\\ &=\frac{r(q-b)(p-a)}{(p-a)(q-b)(r-c)}-\frac{b(p-a)(c-r)}{(p-a)(q-b)(r-c)}-\frac{a(c-r)(q-b)}{(p-a)(q-b)(r-c)}=0 \end{aligned}

\begin{aligned} &=\frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}=0 \\ &=\frac{r}{r-c}+\frac{b+q-q}{q-b}+\frac{a+p-p}{p-a}=0 \\ &=\frac{r}{r-c}+\frac{b-q}{q-b}+\frac{q}{q-b}+\frac{a-p}{p-a}-\frac{p}{p-a}=0 \\ &=\frac{r}{r-c}-1+\frac{q}{q-b}-1-\frac{p}{p-a}=0 \\ &=\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2 \end{aligned}

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