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(C) The equation of a conic passing through the intersection of lines $L_1, L_2, L_3$ is given by $\lambda_1 L_1 L_2 + \lambda_2 L_2 L_3 + \lambda_3 L

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$3$

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(C) The equation of a conic passing through the intersection of lines $L_1, L_2, L_3$ is given by $\lambda_1 L_1 L_2 + \lambda_2 L_2 L_3 + \lambda_3 L_3 L_1 = 0$. For this to represent a circle,the coefficient of $x^2$ must equal the coefficient of $y^2$,and the coefficient of $xy$ must be $0$. $L_1 = x+y$,$L_2 = 2x+y-1$,$L_3 = x-3y+2$. Substituting these,the equation is $\lambda_1(x+y)(2x+y-1) + \lambda_2(2x+y-1)(x-3y+2) + \lambda_3(x-3y+2)(x+y) = 0$. Expanding the terms: $x^2$ coefficient: $2\lambda_1 + 2\lambda_2 + \lambda_3 = C$ $y^2$ coefficient: $\lambda_1 - 3\lambda_2 - 3\lambda_3 = C$ $xy$ coefficient: $3\lambda_1 - 5\lambda_2 - 2\lambda_3 = 0$. From $xy$ coefficient: $2\lambda_3 = 3\lambda_1 - 5\lambda_2$. Equating $x^2$ and $y^2$ coefficients: $2\lambda_1 + 2\lambda_2 + \lambda_3 = \lambda_1 - 3\lambda_2 - 3\lambda_3 \implies \lambda_1 + 5\lambda_2 + 4\lambda_3 = 0$. Substituting $\lambda_3 = \frac{3\lambda_1 - 5\lambda_2}{2}$: $\lambda_1 + 5\lambda_2 + 4(\frac{3\lambda_1 - 5\lambda_2}{2}) = 0 \lambda_1 + 5\lambda_2 + 6\lambda_1 - 10\lambda_2 = 0 7\lambda_1 - 5\lambda_2 = 0 \implies \frac{\lambda_1}{\lambda_2} = \frac{5}{7}$. Then $\lambda_3 = \frac{3(5) - 5(7)}{2} = \frac{15-35}{2} = -10$. So $\frac{\lambda_3}{\lambda_1} = \frac{-10}{5} = -2$. Thus $\frac{7\lambda_1}{\lambda_2} + \frac{\lambda_3}{\lambda_1} = 7(\frac{5}{7}) + (-2) = 5 - 2 = 3$.

If the equation of the circumcircle of the triangle formed by the lines $L_1 \equiv x+y=0$,$L_2 \equiv 2x+y-1=0$,and $L_3 \equiv x-3y+2=0$ is $\lambda_1 L_1 L_2 + \lambda_2 L_2 L_3 + \lambda_3 L_3 L_1 = 0$,then find the value of $\frac{7 \lambda_1}{\lambda_2} + \frac{\lambda_3}{\lambda_1}$.

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