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. $\sqrt{2x}+y=3\sqrt{2}−1$
• C. $x+y+z=\sqrt{3}$
• D. $x−\sqrt{2y}=1−\sqrt{2}$

Answer 1

Answer: (A)

PLAN
(i) Equation of plane through intersection of two planes,
$i.e,(a_{1}x+b_{1}y+c_{1}z+d_1)+λ$
$(a_{2}x+b_{2}y+c_{2}z+d_2)=0$
(ii) Distance of a point $(x_1,y_1,z_1)$ from
$ax+by+cz+d=0$
$=\frac{âˆ\pounds a{x}_1+b{y}_1+c{z}_1+dâˆ\pounds }{\sqrt{a^2+b^2+c^2}}$
Equation of plane passing through intersection of two
planes $x+2y+3z=2$ and $x−y+z=3$ is
$(x+2y+3z−2)+λ(x−y+z−3)=0$
$⇒(1+λ)x+(2+λ)y+(3+λ)z−(2+3λ)=0$
whose distance from $(3,1,−1)$ is $\frac{2}{\sqrt{3}}.$
$⇒\frac{âˆ\pounds 3(1+λ)+1.(2−λ)−1(3+λ)−(2+3λ)âˆ\pounds }{\sqrt{(1+λ{)}^2+(2−λ{)}^2+(3+λ{)}^2}}=\frac{2}{\sqrt{3}}$
$⇒$ $\frac{âˆ\pounds −2λâˆ\pounds }{\sqrt{3{λ}^2+4{λ}^2+14}}=\frac{2}{\sqrt{3}}$
$⇒3{λ}^2=3{λ}^2+4λ+14$
$⇒λ=−\frac{7}{2}$
$∴(1−\frac{7}{2})x+(2−\frac{7}{2})y+(3−\frac{7}{2})z−(2−\frac{21}{2})=0$
$⇒$ $−\frac{5x}{2}+\frac{11}{2}y−\frac{1}{2}z+\frac{17}{2}=0$
or $5x−11y+z+17=0$