# Equation of a plane at a distance

Question:

Equation of a plane at a distance $\sqrt{\frac{2}{21}}$ from the origin, which contains the line of intersection of the planes $x-y-z-1=0$ and $2 x+y-3 z+4=0$, is :

1. $3 x-y-5 z+2=0$

2. $3 x-4 z+3=0$

3. $-x+2 y+2 z-3=0$

4. $4 x-y-5 z+2=0$

Correct Option: , 4

Solution:

Required equation of plane

$\mathrm{P}_{1}+\lambda \mathrm{P}_{2}=0$

$(x-y-z-1)+\lambda(2 x+y-3 z+4)=0$

Given that its dist. From origin is $\frac{2}{\sqrt{21}}$

Thus $\frac{|4 \lambda-1|}{\sqrt{(2 \lambda+1)^{2}+(\lambda-1)^{2}+(-3 \lambda-1)^{2}}}=\frac{\sqrt{2}}{\sqrt{21}}$

$\Rightarrow 21(4 \lambda-1)^{2}=2\left(14 \lambda^{2}+8 \lambda+3\right)$

$\Rightarrow 336 \lambda^{2}-168 \lambda+21=28 \lambda^{2}+16 \lambda+6$

$\Rightarrow 308 \lambda^{2}-184 \lambda+15=0$

$\Rightarrow 308 \lambda^{2}-154 \lambda-30 \lambda+15=0$

$\Rightarrow(2 \lambda-1)(154 \lambda-15)=0$

$\Rightarrow \lambda=\frac{1}{2}$ or $\frac{15}{154}$

for $\lambda=\frac{1}{2}$ reqd. plane is

$4 x-y-5 z+2=0$