If $$\alpha \neq a, \beta \neq b, \gamma \neq c$$ and $$\begin{vmatrix} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{vmatrix} = 0$$, then $$\frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c}$$ is equal to :

Expanding the given determinant or performing row operations ($$R_2 \to R_2-R_1, R_3 \to R_3-R_1$$) and dividing the resulting equation by $$(\alpha-a)(\beta-b)(\gamma-c)$$ yields:

$$\frac{\alpha}{\alpha-a} + \frac{b}{\beta-b} + \frac{c}{\gamma-c} = 0$$

Rewrite the first term as $$1 + \frac{a}{\alpha-a}$$:

$$\frac{a}{\alpha-a} + \frac{b}{\beta-b} + \frac{c}{\gamma-c} = -1$$

The required expression uses $$\gamma$$ instead of $$c$$ in the final numerator. Rewrite $$\frac{\gamma}{\gamma-c}$$ as $$1 + \frac{c}{\gamma-c}$$:

$$\frac{a}{\alpha-a} + \frac{b}{\beta-b} + \left(1 + \frac{c}{\gamma-c}\right) = 1 + (-1) = 0$$