Question

Find the equation of the plane containing the line $2x-y+z-3=0,3x+y+z=5$ and at
a distance of $\frac{1}{\sqrt{6}}$ from the point $(2,1,-1)$

Answer 1

Let the equation of plane be $(3\lambda +2)x+(\lambda -1)y+(y+1)z-5\lambda -3=0$

$\Rightarrow \left|\frac{6\lambda +4+\lambda -1-\lambda -1-5\lambda -3}{{\sqrt{(3\lambda +2)}}^2+(\lambda -1{)}^2+(\lambda +1{)}^2}\right|$
$=\frac{1}{\sqrt{6}}\Rightarrow \left|\frac{6\lambda +4+\lambda -1-\lambda -1-5\lambda -3}{{\sqrt{(3\lambda +2)}}^2+(\lambda -1{)}^2+(\lambda +1{)}^2}\right|=\frac{1}{\sqrt{6}}$ $\Rightarrow 6(\lambda -1{)}^2=11{\lambda }^2+12\lambda +6$

$\Rightarrow \lambda =0,-\frac{24}{5}.$