Question

The equation of the plane through the line $x+y+z+3=0=2x-y+3z+1$ and parallel to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ is

• A. $x-5y+3z=7$
• B. $x-5y+3z=-7$
• C. $x+5y+3z=7$
• D. $x+5y+3z=-0$

Answer 1

The correct option is A $x-5y+3z=7$

Given lines are $x+y+z+3=0=2x-y+3z+1$

Plane equation is

$x+y+z+3+\lambda (2x-y+3z+1)=0(1+2\lambda )x+(1-\lambda )y+(1+3\lambda )z+3+\lambda =0\left(1\right)$

Parallel to $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\left(2\right)$

Normal line of plane $\left(1\right)$ has d.r's $1+2\lambda ,1-\lambda ,1+3\lambda$ are perpendicular to line $\left(2\right)$

By perpendicular property

$(1+2\lambda )+2(1-\lambda )+3(1+3\lambda )=01+2\lambda +2-2\lambda +3+9\lambda =0-6=9\lambda \lambda =\frac{-2}{3}$

Put $\lambda =\frac{-2}{3}$ in equation $\left(1\right)$

$x+y+z+3-\frac{2}{3}(2x-y+3z+1)=0-x+5y-3z=-7x-5y+3z=7$

Equation of plane is $x-5y+3z=7$