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

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

Equation of the required plane is

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

$\Rightarrow (1+K)x+(2-K)y+(3+K)z-(2+3K)=0$

Its distance from $(3,1,-1)$ is $\frac{2}{\sqrt{3}}$

$\Rightarrow \frac{\mid 3(1+K)+(2-K)-(3+K)-(2+3K)\mid }{\sqrt{(K+1{)}^2+(2-K{)}^2+(3+K{)}^2}}$ $=\frac{2}{\sqrt{3}}$

$\Rightarrow \frac{4}{3}=\frac{(-2K{)}^2}{2K^2+4K-14}\Rightarrow 3K^2+4K+14=3K^2$

$\Rightarrow K=\frac{-7}{2}\Rightarrow \frac{-5}{2}x+\frac{11}{2}y-\frac{z}{2}+\frac{17}{2}=0$

$\Rightarrow -5x+11y-z+17=0$

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