Let the equation of plane passing through the line of intersection of the planes x+2 y+a z=2 and x-y+z=3 be 5 x-11 y+b z=6 a-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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Answer: D. -4 Given the equation of the plane passing through the intersection of the two given planes: P: (x + 2y + az - 2) + \lambda(x - y + z - 3) = 0 \Rightarrow x(\lambda+1)+y(2-\lambda)+z(a+\lambda)-2-3 \lambda=0 This is the same as the given equation 5x - 11y + bz = 6a - 1. Now, comparing the coefficients of the corresponding variables in both equations: \frac{\lambda+1}{5} = \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} = \frac{2+3\lambda}{6a-1} Solving for \lambda: -11\lambda -11 = 10 - 5\lambda 6\lambda = -21 \Rightarrow \lambda = -\frac{7}{2} Now, substituting the value of \lambda back into the equations: \frac{2-\lambda}{-11} = \frac{2+3\lambda}{6a-1} \Rightarrow \frac{2+\frac{7}{2}}{-11} = \frac{2-\frac{21}{2}}{6a-1} From this equation, we find the value of a : 6a - 1 = 17 \Rightarrow a = 3 Now, substituting the value of a and \lambda into the equation: \frac{2-\lambda}{-11} = \frac{a+\lambda}{b} \Rightarrow -\frac{1}{2} = \frac{3 - \frac{7}{2}}{b} \Rightarrow -\frac{b}{2} = -\frac{1}{2} \Rightarrow b = 1 Therefore, the point (a, -c, c) \equiv (3, -c, c). The given distance is \frac{2}{\sqrt{a}} = \frac{2}{\sqrt{3}}. The plane is: 5x - 11y + z = 17. Now, let's find the distance: \left|\frac{15 + 11c + c - 17}{\sqrt{147}}\right| = \frac{2}{\sqrt{3}} \Rightarrow c = -1, \frac{4}{3} Since c \in \mathbb{Z}, we have c = -1. Therefore, \frac{a+b}{c} = \frac{3+1}{-1} = -4.

Let the equation of plane passing through the line of intersection of the planes $$x+2 y+a z=2$$ and $$x-y+z=3$$ be $$5 x-11 y+b z=6 a-1$$. 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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