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(C) The equation of the plane passing through the intersection of $x+2y+az-2=0$ and $x-y+z-3=0$ is given by $(x+2y+az-2) + \lambda(x-y+z-3) = 0$.This

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$-4$

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(C) The equation of the plane passing through the intersection of $x+2y+az-2=0$ and $x-y+z-3=0$ is given by $(x+2y+az-2) + \lambda(x-y+z-3) = 0$.This simplifies to $(1+\lambda)x + (2-\lambda)y + (a+\lambda)z - (2+3\lambda) = 0$.Comparing this with the given plane $5x-11y+bz = 6a-1$,we have:$\frac{1+\lambda}{5} = \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} = \frac{2+3\lambda}{6a-1}$.From $\frac{1+\lambda}{5} = \frac{2-\lambda}{-11}$,we get $-11-11\lambda = 10-5\lambda$,so $-6\lambda = 21$,which gives $\lambda = -\frac{7}{2}$.Substituting $\lambda = -\frac{7}{2}$ into the ratios:$\frac{1-3.5}{5} = -0.5$,so $\frac{2+3(-3.5)}{6a-1} = -0.5 \implies \frac{2-10.5}{6a-1} = -0.5 \implies \frac{-8.5}{6a-1} = -0.5 \implies 6a-1 = 17 \implies 6a = 18 \implies a = 3$.Now,$\frac{a+\lambda}{b} = -0.5 \implies \frac{3-3.5}{b} = -0.5 \implies \frac{-0.5}{b} = -0.5 \implies b = 1$.The plane is $5x-11y+z = 17$.The distance from $(a, -c, c) = (3, -c, c)$ to $5x-11y+z-17=0$ is $\frac{|5(3)-11(-c)+c-17|}{\sqrt{5^2+(-11)^2+1^2}} = \frac{|15+11c+c-17|}{\sqrt{25+121+1}} = \frac{|12c-2|}{\sqrt{147}}$.Given distance is $\frac{2}{\sqrt{a}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{49}}{\sqrt{147}} = \frac{14}{\sqrt{147}}$.So,$|12c-2| = 14$. This gives $12c-2 = 14 \implies 12c = 16$ (no integer solution) or $12c-2 = -14 \implies 12c = -12 \implies c = -1$.Thus,$\frac{a+b}{c} = \frac{3+1}{-1} = -4$.

Let the equation of the plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1$. For $c \in \mathbb{Z}$,if the distance of this plane from the point $(a, -c, c)$ is $\frac{2}{\sqrt{a}}$,then $\frac{a+b}{c}$ is equal to

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