Let the plane P: $4 x-y+z=10$ 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 $\mathrm{P}$, then $35 \alpha$ is equal to :

Equation of plane after rotation : <br/><br/>$$ \begin{aligned} & (4 x-y+z-10)+\lambda(x+y-z-y)=0 \\\\ \Rightarrow & (4+\lambda) x+y(\lambda-1)+z(1-\lambda)-4 \lambda-10=0 \\\\ & \overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 \\\\ \Rightarrow & (4+\lambda) 4+(\lambda-1)(-1)+(1-\lambda) 1=0 \\\\ \Rightarrow & 16+4 \lambda-\lambda+1+1-\lambda=0 \\\\ \Rightarrow & 2 \lambda=-18 \\\\ \Rightarrow & \lambda=-9 \end{aligned} $$ <br/><br/>$\therefore$ equation of plane : $-5 x-10 y+10 z+26=0$ <br/><br/>Distance of plane from $(2,3,-4)$ <br/><br/>$$ \begin{aligned} & =\left|\frac{-10-30-40+26}{\sqrt{100+100+26}}\right|=\frac{54}{15}=\alpha \\\\ \therefore 35 \alpha & =35 \cdot \frac{54}{15}=7 \times \frac{54}{3}=7 \times 18=126 \end{aligned} $$