Let the equation of 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 $c\in \mathrm{\mathbb{Z}},$ if the distance of this plane from the point $\left(a,-c,c\right)$ is $\frac{2}{\sqrt{a}},$ then $\frac{a+b}{c}$ is equal to

Let the equation of plane passing through the intersection of two given planes:

$x+2y+az-2+\lambda (x-y+z-3)=0$

$\Rightarrow x(\lambda +1)+y(2-\lambda )+z(a+\lambda )-2-3\lambda =0$

This is same as $5x-11y+bz=6a-1$

$\frac{\lambda +1}{5}=\frac{2-\lambda }{-11}=\frac{a+\lambda }{b}=\frac{2+3\lambda }{6a-1}$

$\Rightarrow -11\lambda -11=10-5\lambda$

$\Rightarrow 6\lambda =-21\Rightarrow \lambda =\frac{-7}{2}$

$\frac{2-\lambda }{-11}=\frac{2+3\lambda }{6a-1}\Rightarrow \frac{2+\frac{7}{2}}{-11}=\frac{2-\frac{21}{2}}{6a-1}$$\Rightarrow -\frac{1}{2}=\frac{-17}{2(6a-1)}\Rightarrow 6a-1=17\Rightarrow a=3$

$\frac{2-\lambda }{-11}=\frac{a+\lambda }{b}\Rightarrow -\frac{1}{2}=\frac{3-\frac{7}{2}}{b}$

$\Rightarrow -\frac{b}{2}=-\frac{1}{2}\Rightarrow b=1$

$\therefore$ point $(a,-c,c)\equiv (3,-c,c)$

Distance$=\frac{2}{\sqrt{a}}=\frac{2}{\sqrt{3}}$

Plane$:5 x-11 y+z=17$

The distance of the plane from $(3,-c,c)$ is

$\left|\frac{15+11c+c-17}{\sqrt{147}}\right|=\frac{2}{\sqrt{3}}$

$\Rightarrow |12c-2|=14$

$\Rightarrow c=-1,\frac{4}{3}$

$c=-1 \left(c\in \mathrm{\mathbb{Z}}\right)$

$\therefore \frac{a+b}{c}=\frac{3+1}{-1}=-4$

Hence this is the required option.