The equation of a plane passing through the line of intersection of the planes $x+2 y+3 z=2$ and $x-y+z=3$…
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
$5 x-11 y+z=17$
$\sqrt{2} x+y=3 \sqrt{2}-1$
$x+y+z=\sqrt{3}$
$x-\sqrt{2} y=1-\sqrt{2}$
Solution
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 $5 x-11 y+z=17$