Question

The equation of line 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. $x-\sqrt{2y}=1-\sqrt{2}$

Answer 1

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

Equation of plane passing through the

line line of intersections the

plane given is given by

$R(x+2y+3z-2)+(x-y+z-3)=0$

i.e $(k+1)x+(2k-1)y+(3k+1)-z(2k+3)=0$

Since, the perpendicular distance from (3,1,-1) is

given by $2/\sqrt{3},$ we get

$\frac{3k+3+2k-1-3k-1-2k-3}{\sqrt{14k^2+14k+3}}=\frac{2}{\sqrt{3}}$

$\Rightarrow k(7k+2)=0\Rightarrow k=0$ or $\frac{-2}{7}$

Hence $e{q}^n$ of plane is $x-y+z=3$

or $5x-11y+z-17=0$