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 from the point $(3,1,–1)$ is

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

Answer 1

The correct option is D $5x–11y+z=17$

Required equation of plane is given by

$P1+\lambda P2=0$

$\Rightarrow (x+2y+3z–2)+\lambda (x–y+z–3)=0$

$\Rightarrow (1+\lambda )x+(2–\lambda )y+(3+\lambda )z–(2+3\lambda )=0...\left(i\right)$

As, $\left|\frac{1(1+\lambda )(2-\lambda )-(3+\lambda )-(2+3\lambda )}{\sqrt{(1+\lambda {)}^2}+(2-\lambda {)}^2+(3+\lambda {)}^2}\right|=\frac{2}{\sqrt{3}}$

$\Rightarrow \lambda =-\frac{7}{2}$

So, using (i), equation of plane becomes

$\Rightarrow 5x-11y+z=17$