The equation of plane passing through the intersection of planes $$x+2 y+3 z-2=0$$ and $$x-y+z-3=0$$ is

$$\begin{aligned} & \Rightarrow(x+2 y+3 z-2)+\lambda(x-y+z-3)=0 \\ & \Rightarrow(\lambda+1) x+(-\lambda+2) y+(\lambda+3) z-(3 \lambda+2)=0 \quad \text{... (i)} \end{aligned}$$

$$\text { Given, the distance from }(3,1,-1) \text { is } \frac{2}{\sqrt{3}}$$

$$\begin{aligned} & \Rightarrow \frac{2}{\sqrt{3}}=\frac{|3(\lambda+1)+1 \cdot(-\lambda+2)-1 \cdot(\lambda+3)-(3 \lambda+2)|}{\sqrt{(\lambda+1)^2+(-\lambda+2)^2+(\lambda+3)^2}} \\ & \Rightarrow \frac{2}{\sqrt{3}}=\frac{|-2 \lambda|}{\sqrt{3 \lambda^2+4 \lambda+14}} \\ & \Rightarrow 2 \sqrt{3 \lambda^2+4 \lambda+14}=2|\lambda| \sqrt{3} \\ & \Rightarrow \sqrt{3 x^2+4 \lambda+14}=\sqrt{3}|\lambda| \end{aligned}$$

On squaring both side

$$\begin{aligned} & \Rightarrow 3 \lambda^2+4 \lambda+14=3 \lambda^2 \\ & \Rightarrow \lambda=-\frac{7}{2} \end{aligned}$$

Put $$\lambda=-\frac{7}{2}$$ in the equation (i)

$$\begin{aligned} & \Rightarrow\left(-\frac{7}{2}+1\right) x+\left(\frac{7}{2}+2\right) y+\left(-\frac{7}{2}+3\right) z +\frac{21}{2}-2=0 \\ & \Rightarrow \frac{-5}{2} x+\frac{11}{2} y-\frac{1}{2} z+\frac{17}{2}=0 \\ & \Rightarrow \quad 5 x-11 y+z=17 \\ \end{aligned}$$