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. $x-\sqrt{2}y=1-2\sqrt{2}$

Answer 1

The correct option is A

$5x-11y+z=17$

(i) equation of a plane through intersection of two planes
$(a_{1}x+b_{1}y+c_{1}z+d_1)+\lambda (a_{2}x+b_{2}y+c_{2}z+d_2)=0$
(ii) distance of a point $(x_1,y_1,z_1)$ from
$ax+by+cz+d=0\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$
Equation of plane passing through intersection of two planes
$x+2y+3z=2$ and $x-y+z=3$ is
$(x+2y+3z-2)+\lambda (x-y+z-3)=0\Rightarrow (1+\lambda )x+(2-\lambda )y+(3+\lambda )z-(2+3\lambda )=0$
Whose distance from (3,1,-1) is $\frac{2}{\sqrt{3}}.$
$\Rightarrow \frac{|3(1+\lambda )+1.(2-\lambda )-1(3+\lambda )-(2+3\lambda )|}{\sqrt{(}1+\lambda {)}^2+(2-\lambda {)}^2+(3+\lambda {)}^2}=\frac{2}{\sqrt{3}}\Rightarrow \frac{|-2\lambda |}{\sqrt{3{\lambda }^2+4\lambda +14}}=\frac{2}{\sqrt{3}}\Rightarrow 3{\lambda }^2=3{\lambda }^2+4\lambda +14\Rightarrow \lambda =-\frac{7}{2}\therefore (1-\frac{7}{2})x+(2+\frac{7}{2})y+(3-\frac{7}{2})z-(2-\frac{21}{2})=0$
$\Rightarrow -\frac{5x}{2}+\frac{11}{2}y-\frac{1}{2}z+\frac{17}{2}=0$
or $5x-11y+z-17=0$