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# If $a \ne p,b \ne q,c \ne r$ and$\left| {\begin{array}{*{20}{c}} p&b&c \\ a&q&c \\ a&b&r \end{array}} \right| = 0$ then find the value of $\dfrac{p}{{p - a}} + \dfrac{q}{{q - b}} + \dfrac{r}{{r - c}}$ ?  Hint-In this question first analyse the question, then perform certain steps to simplify the determinant so that solving it is an easy task. Then by simplifying the determinant step by step we can reach the desired answer. Also study carefully the conditions given as they will be useful.

Considering the given condition in the question we can apply some operation on the determinant

Subtract third row from the first and save answer in first i.e.${R_1} \Rightarrow {R_1} - {R_3}$ and also apply ${R_2} \Rightarrow {R_2} - {R_3}{\text{ , where }}{R_1},{R_2}{\text{ and }}{R_3}{\text{ are first, second and third rows}}{\text{.}}$

Therefore, we get$\left| {\begin{array}{*{20}{c}} {{\text{p - a}}}&{{\text{b - q}}}&{{\text{c - r}}} \\ 0&{{\text{q - b}}}&{{\text{c - r}}} \\ 0&{\text{b}}&{\text{r}} \end{array}} \right| = 0$

As coefficient of ${\text{b - q }}$is $0$ .

Solving it further will generate equation

$\left( {{\text{p - a}}} \right)\left| {\begin{array}{*{20}{c}} {{\text{q - b}}}&{{\text{c - r}}} \\ {\text{b}}&{\text{r}} \end{array}} \right| + \left( {{\text{c - r}}} \right)\left| {\begin{array}{*{20}{c}} 0&{{\text{q - b}}} \\ {\text{a}}&{\text{b}} \end{array}} \right| = 0$

Simplifying the above equation, $\left( {{\text{p - a}}} \right)\left[ {{\text{r}}\left( {{\text{q - b}}} \right) - {\text{b}}\left( {{\text{c - r}}} \right)} \right] + \left( {{\text{c - r}}} \right)\left[ { - {\text{a}}\left( {{\text{q - b}}} \right)} \right] = 0$

From given conditions we can deduce that ${\text{p - a}} \ne {\text{0,q - b}} \ne {\text{0,r - c}} \ne {\text{0}}$

Now, by dividing both sides by $\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]$ we get

$\dfrac{{{\text{r}}\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} - \dfrac{{{\text{b}}\left( {{\text{p - a}}} \right)\left( {{\text{c - r}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} - \dfrac{{{\text{a}}\left( {{\text{c - r}}} \right)\left( {{\text{q - b}}} \right)}}{{\left[ {\left( {{\text{p - a}}} \right)\left( {{\text{q - b}}} \right)\left( {{\text{r - c}}} \right)} \right]}} = 0$

By cancelling the same terms in numerator and denominator we get

$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\text{b}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\text{a}}}{{\left( {{\text{p - a}}} \right)}} = 0$

Then we need to add and subtract ${\text{q}}$ from the numerator of term

$\dfrac{{\text{b}}}{{{\text{q - b}}}}$ and add and subtract ${\text{p}}$ from the numerator of term $\dfrac{{\text{a}}}{{{\text{p - a}}}}$

$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{{\text{b + q - q}}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{{\text{a + p - p}}}}{{\left( {{\text{p - a}}} \right)}} = 0$

From this we get

$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\left( {{\text{b - q}}} \right)}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\left( {{\text{a - p}}} \right)}}{{\left( {{\text{p - a}}} \right)}} + \dfrac{{\text{a}}}{{\left( {{\text{p - a}}} \right)}} = 0$

This implies that

$\dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} - 1 + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} - 1 + \dfrac{{\text{p}}}{{\left( {{\text{p - a}}} \right)}} = 0 \\ \dfrac{{\text{r}}}{{\left( {{\text{r - c}}} \right)}} + \dfrac{{\text{q}}}{{\left( {{\text{q - b}}} \right)}} + \dfrac{{\text{p}}}{{\left( {{\text{p - a}}} \right)}} = 2 \\$

Hence the value of the question is 2.

Note- In getting the values of these types of questions we need to know the right steps in solving a determinant and then simplifying it because simplifying a determinant directly can sometimes be confusing and we may not get the desired result. Therefore, for these questions you need to practise solving the determinants.
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