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

Equation of plane containing the line $2x-y+z-3=0,3x+y+z=5$ is $(2x-y+z-3)+\lambda (3x+y+z-5)=0$

$⟹(2+3\lambda )x+(\lambda -1)y+(\lambda +1)z-(3+5\lambda )=0$

Distance of $(2,1,-1)$ from the plane is $\frac{\mid (2+3\lambda )\left(2\right)+(\lambda -1)\left(1\right)+(\lambda +1)(-1)-(3+5\lambda )\mid }{\sqrt{(2+3\lambda {)}^2+(\lambda -1{)}^2+(\lambda +1{)}^2}}=\frac{1}{\sqrt{6}}$

$⟹\frac{\mid \lambda -1\mid }{\sqrt{11{\lambda }^2+12\lambda +6}}=\frac{1}{\sqrt{6}}$

$6{\lambda }^2-12\lambda +6=11{\lambda }^2+12\lambda +6$

$5{\lambda }^2+24\lambda =0⟹\lambda =-\frac{24}{5},0$

Equation of plane is $62x-29y+19z-117=0$ or $2x-y+z-3=0$