If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $$\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$$, then $$\frac{\mathrm{a}}{\alpha-\mathrm{a}}+\frac{\mathrm{b}}{\beta-\mathrm{b}}+\frac{\gamma}{\gamma-\mathrm{c}}$$ is equal to :

<p>$$\left|\begin{array}{lll} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{array}\right|=0$$</p> <p>$$\begin{aligned} & R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\ & \Rightarrow\left|\begin{array}{ccc} \alpha-a & b-\beta & 0 \\ 0 & \beta-b & c-\gamma \\ a & b & \gamma \end{array}\right|=0 \end{aligned}$$</p> <p>Take $\alpha$-a, $\beta$-b, $\gamma$-c common from column-1, 2 and 3 respectively</p> <p>$$\begin{aligned} & (\alpha-a)(\beta-b)(\gamma-c)\left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \frac{a}{\alpha-a} & \frac{b}{\beta-b} & \frac{\gamma}{\gamma-c} \end{array}\right|=0 \\ & \Rightarrow \frac{\gamma}{\gamma-c}+\frac{b}{\beta-b}+\frac{a}{\alpha-a}=0 \end{aligned}$$</p>