Let the plane P: 4x - y + z = 10 be rotated by an angle (π)/(2)...

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$</p><p>$⇒ \left(4 + \lambda\right)x + y\left(\lambda - 1\right) + z\left(1 - \lambda\right) - 4\lambda - 10 = 0$</p><p>Now angle between the new plane and$4x - y + z - 10$is$90 °$.</p><p>$⇒ \overset{\to}{n}1 \cdot \overset{\to}{n}2 = 0$</p><p>$⇒ \left(4 + \lambda\right)4 + \left(\lambda - 1\right)\left(- 1\right) + \left(1 - \lambda\right)1 = 0$</p><p>$⇒ 16 + 4\lambda - \lambda + 1 + 1 - \lambda = 0$</p><p>$⇒ 2\lambda = - 18$</p><p>$⇒ \lambda = - 9$</p><p>$∴$Equation of plane :$– 5x – 10y + 10z + 26 = 0$</p><p>Distance of plane from$( 2, 3, – 4 )$</p><p>$= \left|\dfrac{- 10 - 30 - 40 + 26}{\sqrt{25 + 100 + 100}}\right| = \dfrac{54}{15} = \alpha$</p><p>$35\alpha = 35 \cdot \dfrac{54}{15} = 7 \times \dfrac{54}{3} = 7 \times 18 = 126

Hence this is the required option.