# Show that

Question:

If $\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|=0$, find the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}, p \neq a, q \neq b, r \neq c$

Solution:

Let $\Delta=\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|$.

Now,

$\Delta=\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|$

$=\left|\begin{array}{ccc}p & b & c \\ 0 & q-b & c-r \\ a & b & r\end{array}\right|$         $\left[\right.$ Applying $\left.R_{2} \rightarrow R_{2}-R_{3}\right]$

$=p[r(q-b)-b(c-r)]+a[b(c-r)-c(q-b)]$       [Expanding along first column]

$=\operatorname{pr}(q-b)+p b(r-c)-a b(r-c)-a c(q-b)$

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

Since, $\Delta=0$.

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

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

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

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

$\Rightarrow \frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=0$

$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{p}{p-a}+\frac{q}{q-b}-\frac{a}{p-a}-\frac{b}{q-b}$

$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{p-a}{p-a}+\frac{q-b}{q-b}$

$\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2$

Hence, the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is 2 .