Given two planes are $x+2y+3z-2=0$..........(i) and $x-y+z-3=0$.......(ii)

General eqn. of the plane passing through the line of intersection of (i) and (ii) is

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

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

Given that the plane (iii) is at a distance of $\large\frac{2}{\sqrt 3}$ from $(3,1,-1)$.

$\therefore\:\bigg|\large\frac{3(1+\lambda)+(2-\lambda)-(3+\lambda)+(-2-3\lambda)}{\sqrt {(1+\lambda)^2+(2-\lambda)^2+(3+\lambda)^2}}\bigg|=\frac{2}{\sqrt 3}$

$\Rightarrow\:\bigg|\large\frac{-2\lambda}{\sqrt {(1+\lambda)^2+(2-\lambda)^2+(3+\lambda)^2}}\bigg|=\frac{2}{\sqrt 3}$

$\Rightarrow\:12\lambda^2=4(1+\lambda^2+2\lambda+4+\lambda^2-4\lambda+9+\lambda^2+6\lambda)$

$\Rightarrow\:16\lambda+56=0$

$\Rightarrow\:\lambda=-\large\frac{7}{2}$

Substituting the value of $\lambda$ in (iii) the required eqn. of the plane is

$5x-11y+z=17$