Question

The plane denoted by ${\pi }_1:4x+7y+4z+81=0$ is rotated through a right angle about its line of intersection with the plane ${\pi }_2:5x+3y+10z=25$. If the plane in its new position be denoted by $\pi ,$ and the distance of this plane from the origin is $\sqrt{k},$ where $k\in N,$ the $k=$

• A. $212$
• B. $106$
• C. $424$
• D. None of these

Answer 1

The correct option is A $212$

$4x+7y+4z+81=0$ ...(1)

$5x+3y+10z=25$ ...(2)

Equation of plane passing through intersection of (1) and (2) is given by,

$(4x+7y+4z+81)+\lambda (5x+3y+10z-25)=0$

$\Rightarrow (4+5\lambda )x+(7+3\lambda )y+(4+10\lambda )z+81-25\lambda =0$ ...(3)

Plane (2) is perpendicular to (1)

$\Rightarrow 4(4+5\lambda )+7(7+3\lambda )+4(4+10\lambda )=0$

$\Rightarrow \lambda =-1$ putting $\lambda =-1$ in (3), we get

$-x+4y-6z+106=0$ ...(4)

$\because$ distance of plane (4) from $(0,0,0)=\sqrt{k}$(given)

$\Rightarrow \frac{106}{\sqrt{1+16+36}}=\sqrt{k}$

$\Rightarrow k=212$