$If p \neq a, q \neq b, r \neq c$ and the system of equation $p x + a y + a z = 0$ , $b x + q y + b z = 0$ , $c x + c y + r z = 0$ has a non-trival solution, then the value of $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}$ is

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answer is 2.

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Detailed Solution

Given That

$|\begin{matrix} p & a & a \\ b & q & b \\ c & c & r \end{matrix}| = 0$ $\begin{array}{l} c_{2} \to c_{2} - c_{1} \\ c_{3} \to c_{3} - c_{2} \end{array}$

$\Rightarrow |\begin{matrix} p & a - p & a - p \\ b & q - b & 0 \\ c & 0 & r - c \end{matrix}| = 0$

Expanding along c₃, we get

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

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

$= \frac{q - b}{q - b} + \frac{r - c}{r - c} = 1 + 1 = 2$

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