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(A) The equation of any plane passing through the line of intersection of the planes $x+2y+3z-2=0$ and $x-y+z-3=0$ is given by $(x+2y+3z-2) + \lambda(

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$5x-11y+z=17$

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(A) The equation of any plane passing through the line of intersection of the planes $x+2y+3z-2=0$ and $x-y+z-3=0$ is given by $(x+2y+3z-2) + \lambda(x-y+z-3) = 0$.This simplifies to $(1+\lambda)x + (2-\lambda)y + (3+\lambda)z - (2+3\lambda) = 0$.The distance of this plane from the point $(3,1,-1)$ is given by $\frac{|(1+\lambda)(3) + (2-\lambda)(1) + (3+\lambda)(-1) - (2+3\lambda)|}{\sqrt{(1+\lambda)^2 + (2-\lambda)^2 + (3+\lambda)^2}} = \frac{2}{\sqrt{3}}$.Simplifying the numerator: $|3+3\lambda + 2-\lambda - 3-\lambda - 2-3\lambda| = |-2\lambda|$.Simplifying the denominator: $\sqrt{(1+2\lambda+\lambda^2) + (4-4\lambda+\lambda^2) + (9+6\lambda+\lambda^2)} = \sqrt{3\lambda^2+4\lambda+14}$.Thus,$\frac{|-2\lambda|}{\sqrt{3\lambda^2+4\lambda+14}} = \frac{2}{\sqrt{3}}$.Squaring both sides: $\frac{4\lambda^2}{3\lambda^2+4\lambda+14} = \frac{4}{3}$.$3\lambda^2 = 3\lambda^2+4\lambda+14$,which gives $4\lambda = -14$,so $\lambda = -\frac{7}{2}$.Substituting $\lambda = -\frac{7}{2}$ into the equation: $(1-\frac{7}{2})x + (2+\frac{7}{2})y + (3-\frac{7}{2})z - (2-\frac{21}{2}) = 0$.$-\frac{5}{2}x + \frac{11}{2}y - \frac{1}{2}z + \frac{17}{2} = 0$.Multiplying by $-2$,we get $5x-11y+z-17=0$,or $5x-11y+z=17$.

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

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