Question

The equation of a plane passing through the line of intersection of the planes $x+2 y+3 z=2$ and $x-y+z=$ 3 and at a distance $\frac{2}{\sqrt{3}}$ from the point $(3,1,-1)$ is

• A. $5 x-11 y+z=17$
• 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
$P \equiv(x+2 y+3 z-2)+\lambda(x-y+z-3)=0$
$\Rightarrow (1+\lambda) x+(2-\lambda) y+(3+\lambda) z-(2+3 \lambda)=0$
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} $
$\Rightarrow -\frac{5}{2} x+\frac{11}{2} y-\frac{z}{2}+\frac{17}{2}=0$
$-5 x+11 y-z+17=0$