The equation of a plane passing through the line of intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$ is

$\text{Given: planes } x+2y+3z=2,\quad x-y+z=3$

$\text{point } (3, 1, -1), \quad \text{distance } \dfrac{2}{\sqrt{3}}$

$\text{To find: equation of such a plane}$

$\pi_1: x+2y+3z=2$

$\pi_2: x-y+z=3 \ \Rightarrow\ \text{meet in a line } L$

$\text{Family through } L:\ \pi_1 + \lambda\,\pi_2 = 0$

$\lambda = 0 \Rightarrow \pi_1, \qquad \lambda \to \infty \Rightarrow \pi_2$

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

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

$\vec{n}(\lambda) = (1+\lambda,\ 2-\lambda,\ 3+\lambda)$

$d = \dfrac{\big|\vec{n}\cdot (3,1,-1) - (2+3\lambda)\big|}{|\vec{n}|}$

$\vec{n}\cdot(3,1,-1) = 3(1+\lambda) + (2-\lambda) - (3+\lambda) = 2+\lambda$

$\text{numerator} = (2+\lambda) - (2+3\lambda) = -2\lambda \ \Rightarrow\ 2|\lambda|$

$|\vec{n}|^2 = (1+\lambda)^2 + (2-\lambda)^2 + (3+\lambda)^2 = 3\lambda^2 + 4\lambda + 14$

$\dfrac{2|\lambda|}{\sqrt{3\lambda^2+4\lambda+14}} = \dfrac{2}{\sqrt{3}}$

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

$\lambda = -\dfrac{7}{2}:\ -\dfrac52 x + \dfrac{11}{2} y - \dfrac12 z = -\dfrac{17}{2}$

$5x - 11y + z = 17$

$\dfrac{|15-11-1-17|}{\sqrt{25+121+1}} = \dfrac{14}{7\sqrt{3}} = \dfrac{2}{\sqrt{3}} \ \checkmark$

$\boxed{\text{Answer: option (1)}\quad 5x-11y+z=17}$