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 :
Correct Option: , 4
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
$4 x-y-5 z+2=0$