Answer

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**Hint:**To solve this question, firstly we will use row operations ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$ and ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$. After that, we will expand the determinant along row ${{R}_{1}}$. After that, we will divide the whole equation by (p-a)(q–b)(c–r), and by doing some simplification, we will find the value of $\dfrac{p}{p-a}+\dfrac{b}{q-b}+\dfrac{r}{r-c}$.

**Complete step by step answer:**

Now, before we start solving the questions, let us see how we calculate determinant and what are its various properties

Now , if we want to calculate the determinant of matrix A of order $3\times 3$, then determinant of matrix A of $3\times 3$ is evaluated as,

$\left| \begin{matrix}

{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\

{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\

{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\

\end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$

Some of the properties of determinant are as follows,

( a ) Determinant evaluated across any row or column is the same.

( b ) If an element of a row or a column are zeros, then the value of the determinant is equal to zero.

( c ) If rows and columns are interchanged then the value of the determinant remains the same.

( d ) Determinant of an identity matrix is 1.

Now, let us move to question now, it is asked to find the value of $\dfrac{p}{p-a}+\dfrac{b}{q-b}+\dfrac{r}{r-c}$ and in question it is given that $a\ne p$ , $b\ne q$ and $c\ne r$,and $\left| \begin{matrix}

p & b & c \\

a & q & c \\

a & b & r \\

\end{matrix} \right|=\text{0}$.

Now, using elementary row operation ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$ , we get

$\left| \begin{matrix}

p-a & b-q & c-c \\

a & q & c \\

a & b & r \\

\end{matrix} \right|=\text{0}$

using elementary row operation ${{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$ , we get

$\left| \begin{matrix}

p-a & b-q & 0 \\

0 & q-b & c-r \\

a & b & r \\

\end{matrix} \right|=\text{0}$

Now, expanding determinant along ${{R}_{1}}$, we get

$(p-a)(r(q-b)-b(c-r))-(b-q)(0-a(c-r))+0=0$

On simplifying, we get

$(p-a)(r(q-b)-b(c-r))+(b-q)a(c-r)=0$

$\Rightarrow r(p-a)(q-b)-b(p-a)(c-r)+a(b-q)(c-r)=0$

Now, dividing the whole equation by ( p –a )( q – b )( c – r ), we get

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

On simplifying, we get

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

Or, \[\Rightarrow \dfrac{r}{(c-r)}-\dfrac{b}{(q-b)}-\dfrac{a}{(p-a)}=0\]

Re – writing above equation, we get

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

Multiplying both sides by -1, we get

\[\dfrac{r}{(r-c)}+\dfrac{b}{(q-b)}+\dfrac{a}{(p-a)}=0\]

Now, adding and subtracting q in numerator of term \[\dfrac{b}{(q-b)}\] and adding and subtracting p in numerator of \[\dfrac{a}{(p-a)}\], we get

\[\dfrac{r}{(r-c)}+\dfrac{b+q-q}{(q-b)}+\dfrac{a+p-p}{(p-a)}=0\]

\[\Rightarrow \dfrac{r}{(r-c)}+\dfrac{b-q}{(q-b)}+\dfrac{q}{(q-b)}+\dfrac{a-p}{(p-a)}+\dfrac{p}{(p-a)}=0\]

On simplification, we get

\[\dfrac{r}{(r-c)}-1+\dfrac{q}{(q-b)}-1+\dfrac{p}{(p-a)}=0\]

On solving, we get

\[\dfrac{r}{(r-c)}+\dfrac{q}{(q-b)}+\dfrac{p}{(p-a)}=2\]

**Note:**It is very important to know how to solve determinant using it’s properties so knowledge of properties of determinant should be a priority. Always remember that $\left| \begin{matrix}

{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\

{{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\

{{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\

\end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$. In determinant we can use both column and row elementary transformation. Calculation should be done carefully while solving determinant problems.

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