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

Plane through the line of intersection of the planes be
$(\mathrm{x}+2\mathrm{y}+\mathrm{az}-2)+\mathrm{\lambda }(\mathrm{x}-\mathrm{y}+\mathrm{z}-3)=0\Rightarrow (1+\mathrm{\lambda })\mathrm{x}+(2-\mathrm{\lambda })\mathrm{y}+(\mathrm{a}+\mathrm{\lambda })\mathrm{z}-(2+3\mathrm{\lambda })=0................\left(1\right)$
But the plane is given as $5\mathrm{x}-11\mathrm{y}+6\mathrm{z}=6\mathrm{a}-1.................\left(2\right)$
$\therefore \frac{1+\mathrm{\lambda }}{5}=\frac{2-\mathrm{\lambda }}{-11}=\frac{\mathrm{a}+\mathrm{\lambda }}{6}=\frac{2+3\mathrm{\lambda }}{6\mathrm{a}-1}\Rightarrow \mathrm{\lambda }=\frac{-7}{2},\mathrm{a}=3,\mathrm{b}=1$
Plane is $5\mathrm{x}-11\mathrm{y}+\mathrm{z}=17$ 
Distance from $(\mathrm{a},-\mathrm{c},\mathrm{c})=\frac{2}{\sqrt{\mathrm{a}}}$
$\Rightarrow \frac{|15.3+11\mathrm{c}+\mathrm{c}-17|}{\sqrt{5^2+{11}^2+1^2}}=\frac{2}{\sqrt{3}}$ 
$$\begin{gathered} \Rightarrow -2+12\mathrm{c}=\pm 14 \\ \Rightarrow \mathrm{c}=-1\because \mathrm{c}\in \mathrm{z} \end{gathered}$$