If alpha neq a, beta neq b, gamma neq c and l | vedclass

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(C) Given the determinant equation:$\left|\begin{array}{lll}\alpha & b & c \ a & \beta & c \ a & b & \gamma\end{array}ight|=0$Apply row operations

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(C) Given the determinant equation:$\left|\begin{array}{lll}\alpha & b & c \ a & \beta & c \ a & b & \gamma\end{array}ight|=0$Apply row operations $R_1 ightarrow R_1 - R_2$ and $R_2 ightarrow R_2 - R_3$:$\left|\begin{array}{ccc}\alpha-a & b-\beta & 0 \ 0 & \beta-b & c-\gamma \ a & b & \gamma\end{array}ight|=0$Expanding along the first row:$(\alpha-a)[(\beta-b)\gamma - b(c-\gamma)] - (b-\beta)[0 - a(c-\gamma)] + 0 = 0$$(\alpha-a)(\beta-b)\gamma - b(\alpha-a)(c-\gamma) + a(b-\beta)(c-\gamma) = 0$Divide the entire equation by $(\alpha-a)(\beta-b)(\gamma-c)$:$\frac{(\alpha-a)(\beta-b)\gamma}{(\alpha-a)(\beta-b)(\gamma-c)} - \frac{b(\alpha-a)(c-\gamma)}{(\alpha-a)(\beta-b)(\gamma-c)} + \frac{a(b-\beta)(c-\gamma)}{(\alpha-a)(\beta-b)(\gamma-c)} = 0$$\frac{\gamma}{\gamma-c} + \frac{b}{\beta-b} + \frac{a}{\alpha-a} = 0$

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

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