For three consecutive odd integers a, b and c | vedclass

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(C) The given line is $ax + by + c = 0$. Since $a, b,$ and $c$ are three consecutive odd integers,they are in arithmetic progression. Thus,$b - a = c

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$5$

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(C) The given line is $ax + by + c = 0$. Since $a, b,$ and $c$ are three consecutive odd integers,they are in arithmetic progression. Thus,$b - a = c - b$,which implies $c = 2b - a$. Substituting $c$ into the line equation: $ax + by + (2b - a) = 0$. Rearranging the terms: $a(x - 1) + b(y + 2) = 0$. For this line to pass through the fixed point $(\alpha, \beta)$ for all such $a$ and $b$,the coefficients must satisfy: $\alpha - 1 = 0 \Rightarrow \alpha = 1$ and $\beta + 2 = 0 \Rightarrow \beta = -2$. Therefore,the value of $\alpha^2 + \beta^2 = 1^2 + (-2)^2 = 1 + 4 = 5$.

For three consecutive odd integers $a, b$ and $c$,if the variable line $a x+b y+c=0$ always passes through the point $(\alpha, \beta)$,the value of $\alpha^2+\beta^2$ equals

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