Let the plane $\mathrm{P}:4\mathrm{x}-\mathrm{y}+\mathrm{z}=10$ be rotated by an angle $\frac{\mathrm{\pi }}{2}$ about its line of intersection with the plane $\mathrm{x}+\mathrm{y}-\mathrm{z}=4.$ If $\mathrm{\alpha }$ is the distance of the point (2,3,-4) from the new position of the plane P, then $35\mathrm{\alpha }$ is equal to

Equation of the required plane is of the form ${\mathrm{\pi }}_1+{\mathrm{\lambda \pi }}_2=0$
i.e., $4\mathrm{x}-\mathrm{y}+\mathrm{z}-10+\mathrm{\lambda }(\mathrm{x}+\mathrm{y}-\mathrm{z}-4)=0$
$\Rightarrow (4+\mathrm{\lambda })\mathrm{x}+(\mathrm{\lambda }-1)\mathrm{y}+(1-\mathrm{\lambda })\mathrm{z}-(10+4\mathrm{\lambda })=0$
It is perpendicular to  $4\mathrm{x}-\mathrm{y}+\mathrm{z}=10$
$\therefore (4+\mathrm{\lambda })4+(\mathrm{\lambda }-1)(-1)+(1-\mathrm{\lambda })=0\Rightarrow \mathrm{\lambda }=-9$
$\therefore$ Equation of rotated plane is  $-5\mathrm{x}-10\mathrm{y}+10\mathrm{z}+26=0$
$\Rightarrow 5\mathrm{x}+10\mathrm{y}-10\mathrm{z}-26=0$
Perpendicular distance of (2,3,-4) from above plane
$=\left|\frac{10+30+40-26}{\sqrt{25+100+100}}\right|=\frac{18}{5}=\alpha$
$\therefore 35\mathrm{\alpha }=18\times 7=126$