The equation of a plane passing through the line of intersection of the planes \(x+2 y+3 z=2\) and \(x-y+z=3\)

and at a distance \(\frac{2}{\sqrt{3}}\) from the point \((3,1,-1)\) is

Equation of the plane passing through the intersection line of given planes is

\(\begin{array}{ll}

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

\text { or } & (1+\lambda) x+(2-\lambda) y+(3+\lambda) z+(-2-3 \lambda)=0

\end{array}\)

\(\because \quad\) Its distance from the point \((3,1,-1)\) is \(\frac{2}{\sqrt{3}}\)

\(\begin{array}{l}

\therefore\left|\frac{3(1+\lambda)+1(2-\lambda)-1(3+\lambda)+(-2-3 \lambda)}{\sqrt{(1+\lambda)^{2}+(2-\lambda)^{2}+(3+\lambda)^{2}}}\right|=\frac{2}{\sqrt{3}} \\

\Rightarrow\left|\frac{-2 \lambda}{\sqrt{3 \lambda^{2}+4 \lambda+14}}\right|=\frac{2}{\sqrt{3}} \\

\Rightarrow 3 \lambda^{2}+4 \lambda+14=3 \lambda^{2} \Rightarrow \lambda=-\frac{7}{2}

\end{array}\)

\(\therefore\) Required equation of plane is

\((x+2 y+3 z-2)-\frac{7}{2}(x-y+z-3)=0\)

or \(5 x-11 y+z=17\)