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 $2x+y-3z+4=0$, is :

• A. $4x-y-5z+2=0$
• B. $3x-4z+3=0$
• C. $-x+2y+2z-3=0$
• D. $3x-y-5z+2=0$

Answer 1

The correct option is A $4x-y-5z+2=0$

Let the equation of required plane passing through the intersection of the two given planes be,
$(x-y-z-1)+\lambda (2x+y-3z+4)=0$
$\Rightarrow (1+2\lambda )x+(\lambda -1)y-(3\lambda +1)z+(4\lambda -1)=0$
Now, distance from origin is:
$\left|\frac{4\lambda -1}{\sqrt{(2\lambda +1{)}^2+(\lambda -1{)}^2+(3\lambda +1{)}^2}}\right|=\sqrt{\frac{2}{21}}$

$\Rightarrow 21(16{\lambda }^2-8\lambda +1)=2(14{\lambda }^2+8\lambda +3)$
$\Rightarrow 308{\lambda }^2-184\lambda +15=0$
$\Rightarrow 308{\lambda }^2-154\lambda -30\lambda +15=0$
$\Rightarrow (2\lambda -1)(154\lambda -15)=0$
$\lambda =\frac{1}{2}$ or $\frac{15}{154}$
For $\lambda =\frac{1}{2}$ required equation of plane
$2x-\frac{1}{2}y-\frac{5}{2}z+1=0$
$\therefore 4x-y-5z+2=0$