If p ≠ a, q ≠ b, r ≠ c and the system of equations aligned & p x+a...

As the given system of equations has a non-trivial solution. \Delta=\left|$\begin{array}{lll}p & a & a \\ b & q & b \\ c & c & r\end{array}$\right|=0 Applying $C_2 \rightarrow C_2-C_1$ and C_3 \rightarrow C_3-C_1\Delta=\left|$\begin{array}{ccc}p & a-p & a-p \\ b & q-b & 0 \\ c & 0 & r-c\end{array}$\right|=0 Expanding along $C_3$, we get (a-p)\left|$\begin{array}{cc}b & q-b \\ c & 0\end{array}$\right|+(r-c)\left|$\begin{array}{cc}p & a-p \\ b & q-b\end{array}$\right|=0$\begin{aligned} & \Rightarrow(a-p)(-c)(q-b)+(r-c) \\ & \{p(q-b)-b(a-p)\}=0 \\ & \Rightarrow \\ & (p-a)(q-b) c+p(r-c)(q-b)+b(r-c)(p-a)=0 \end{aligned}$$$Dividing by$(p-a)(q-b)(r-c)$, we get$\frac{c}{r-c}+\frac{p}{p-a}+\frac{b}{q-b}=0\Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{q-b}{q-b}+\frac{r-c}{r-c}=2$