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

Answer: (A)

Equation of required plane is
$P\equiv (x+2y+3z-2)+\lambda (x-y+z-3)=0$
$\Rightarrow (1+\lambda )x+(2-\lambda )y+(3+\lambda )z-(2+3\lambda )=0$
Its distance from $(3,1,-1)$ is $\frac{2}{\sqrt{3}}$.
$\Rightarrow \frac{2}{\sqrt{3}}=\frac{|3(1+\lambda )+(2-\lambda )-(3+\lambda )-(2+3\lambda )|}{\sqrt{(\lambda +1{)}^2+(2-\lambda {)}^2+(3+\lambda {)}^2}}$
$=\frac{4}{3}=\frac{(-2\lambda {)}^2}{2{\lambda }^2+4\lambda +14}$
$\Rightarrow 3{\lambda }^2+4\lambda +14=3{\lambda }^2$
$\Rightarrow \lambda =-\frac{7}{2}$
$\Rightarrow -\frac{5}{2}x+\frac{11}{2}y-\frac{z}{2}+\frac{17}{2}=0$
$-5x+11y-z+17=0$