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 :

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