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(B) The equation of a circle passing through the intersection of lines $L_1=0, L_2=0, L_3=0$ is given by $L_1 L_2 + \lambda L_2 L_3 + \mu L_3 L_1 = 0$

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$x^2+y^2-17x-19y+50=0$

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(B) The equation of a circle passing through the intersection of lines $L_1=0, L_2=0, L_3=0$ is given by $L_1 L_2 + \lambda L_2 L_3 + \mu L_3 L_1 = 0$.Let $L_1: x+y-6=0$,$L_2: 2x+y-4=0$,$L_3: x+2y-5=0$.The equation is $(x+y-6)(2x+y-4) + \lambda(2x+y-4)(x+2y-5) + \mu(x+2y-5)(x+y-6) = 0$.Expanding the terms,the coefficient of $x^2$ is $2 + 2\lambda + \mu$ and the coefficient of $y^2$ is $1 + 2\lambda + 2\mu$.For a circle,coeff$(x^2)$ = coeff$(y^2)$ $\Rightarrow 2 + 2\lambda + \mu = 1 + 2\lambda + 2\mu$ $\Rightarrow \mu = 1$.The coefficient of $xy$ is $1 + 2 + 4\lambda + \lambda + \mu + 2\mu = 3 + 5\lambda + 3\mu = 0$.Substituting $\mu = 1$,we get $3 + 5\lambda + 3 = 0$ $\Rightarrow 5\lambda = -6$ $\Rightarrow \lambda = -6/5$.Substituting $\lambda$ and $\mu$ into the equation: $(x+y-6)(2x+y-4) - \frac{6}{5}(2x+y-4)(x+2y-5) + (x+2y-5)(x+y-6) = 0$.Multiplying by $5$: $5(x+y-6)(2x+y-4) - 6(2x+y-4)(x+2y-5) + 5(x+2y-5)(x+y-6) = 0$.Simplifying the expression leads to $x^2+y^2-17x-19y+50=0$.

The equation of the circle circumscribing the triangle formed by the straight lines $x+y=6$,$2x+y=4$,and $x+2y=5$ is given by:

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