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

$\text{Given: } P_1: x + 2y + 3z = 2$ $P_2: x - y + z = 3$ $\text{Distance from } (3,\ 1,\ -1) \text{ to the plane } = \dfrac{2}{\sqrt{3}}$ $\text{To find: equation of the plane through } P_1 \cap P_2$ $\text{at distance } \dfrac{2}{\sqrt{3}} \text{ from } (3,\ 1,\ -1)$ $\text{Required plane contains the whole line } L = P_1 \cap P_2$ $\text{Family: } P_1 + \lambda P_2 = 0$ $P_1 = x+2y+3z-2, \quad P_2 = x-y+z-3$ $(x+2y+3z-2) + \lambda(x-y+z-3) = 0$ $(1+\lambda)x + (2-\lambda)y + (3+\lambda)z - (2+3\lambda) = 0$ $\text{Plan: } 1.\ \text{family plane} = 0$ $2.\ \text{distance from } (3,1,-1) = \dfrac{2}{\sqrt{3}} \Rightarrow \lambda$ $3.\ \text{write the plane, match the options}$ $\dfrac{|A\cdot 3 + B\cdot 1 + C\cdot(-1) + D|}{\sqrt{A^2+B^2+C^2}} = \dfrac{2}{\sqrt{3}}$ $3A + B - C + D = 3(1+\lambda) + (2-\lambda) - (3+\lambda) - (2+3\lambda)$ $= -2\lambda$ $A^2+B^2+C^2 = (1+\lambda)^2 + (2-\lambda)^2 + (3+\lambda)^2$ $= 3\lambda^2 + 4\lambda + 14$ $\dfrac{2|\lambda|}{\sqrt{3\lambda^2+4\lambda+14}} = \dfrac{2}{\sqrt{3}}$ $3\lambda^2 = 3\lambda^2 + 4\lambda + 14 \ \Rightarrow \ \lambda = -\dfrac{7}{2}$ $\left(1+\lambda,\ 2-\lambda,\ 3+\lambda\right) = \left(-\dfrac{5}{2},\ \dfrac{11}{2},\ -\dfrac{1}{2}\right)$ $-(2+3\lambda) = \dfrac{17}{2}$ $-5x + 11y - z + 17 = 0$ $\Rightarrow 5x - 11y + z = 17$ $d = \dfrac{|15 - 11 - 1 - 17|}{\sqrt{25+121+1}} = \dfrac{14}{7\sqrt{3}} = \dfrac{2}{\sqrt{3}} \ \checkmark$ $5x - 11y + z = 17 \quad \Rightarrow \quad \text{Option }(1)$