Question

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

• A. $5x-11y+z=17$
• B. $\sqrt{2}x+y=3\sqrt{2}-1$
• C. $x+y+z=\sqrt{3}$
• D. $x-\sqrt{2}y=1-\sqrt{2}$

Answer 1

The correct option is A $5x-11y+z=17$

Equation of plane passing through intersection of planes $P_1$ and $P_2$ is $P_1+\lambda P_2=0$
$(x+2y+3z-2)+\lambda (x-y+z-3)=0$
$(1+\lambda )x+(2-\lambda )y+(3+\lambda )z-2-3\lambda =0$

Distance of plane from point $(3,1,-1)$ is $\frac{2}{\sqrt{3}}$
$\Rightarrow \frac{2}{\sqrt{3}}=\frac{|3+3\lambda +2-\lambda -3-\lambda -2-3\lambda |}{\sqrt{(1+\lambda {)}^2+(2-\lambda {)}^2+(3+\lambda {)}^2}}$
$\Rightarrow \left|\frac{-2\lambda }{\sqrt{3{\lambda }^2+4\lambda +14}}\right|=\frac{2}{\sqrt{3}}$
$\Rightarrow 3{\lambda }^2=3{\lambda }^2+4\lambda +14$
$\therefore \lambda =-\frac{7}{2}$
So equation of plane is
$(1-\frac{7}{2})x+(2+\frac{7}{2})y+(3-\frac{7}{2})z-2+\frac{21}{2}=0$
$\therefore 5x-11y+z=17$