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

We know that if a plane passes through the line of intersection of planes $P_1$ and $P_2$ then the family of planes is given by $P_1+\lambda P_2$ where $\lambda \in R$.

Equation of plane after rotation :

$\left(4x-y+z-10\right)+\lambda \left(x+y-z-4\right)=0$

$\Rightarrow \left(4+\lambda \right)x+y\left(\lambda -1\right)+z\left(1-\lambda \right)-4\lambda -10=0$

Now angle between the new plane and $4x-y+z-10$ is $90°$.

$\Rightarrow {\overrightarrow{n}}_1·{\overrightarrow{n}}_2=0$

$\Rightarrow \left(4+\lambda \right)4+\left(\lambda -1\right)\left(-1\right)+\left(1-\lambda \right)1=0$

$\Rightarrow 16+4\lambda -\lambda +1+1-\lambda =0$

$\Rightarrow 2\lambda =-18$

$\Rightarrow \lambda =-9$

$\therefore$ Equation of plane : $–5x–10y+10z+26=0$

Distance of plane from $(2, 3, –4)$

$=\left|\frac{-10-30-40+26}{\sqrt{25+100+100}}\right|=\frac{54}{15}=\alpha$

$35\alpha =35·\frac{54}{15}=7\times \frac{54}{3}=7\times 18=126$

Hence this is the required option.