Let the plane P: 4x - y + z = 10 be rotated b | vedclass

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(D) The equation of the family of planes passing through the line of intersection of $P_1: 4x - y + z - 10 = 0$ and $P_2: x + y - z - 4 = 0$ is given

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$126$

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(D) The equation of the family of planes passing through the line of intersection of $P_1: 4x - y + z - 10 = 0$ and $P_2: x + y - z - 4 = 0$ is given by $(4x - y + z - 10) + \lambda(x + y - z - 4) = 0$,which simplifies to $(4 + \lambda)x + (-1 + \lambda)y + (1 - \lambda)z - (10 + 4\lambda) = 0$.The normal vector of the original plane $P$ is $\vec{n}_1 = (4, -1, 1)$ and the normal vector of the new plane is $\vec{n}_2 = (4 + \lambda, -1 + \lambda, 1 - \lambda)$.Since the angle between the planes is $\frac{\pi}{2}$,their normal vectors are perpendicular,so $\vec{n}_1 \cdot \vec{n}_2 = 0$.$4(4 + \lambda) - 1(-1 + \lambda) + 1(1 - \lambda) = 0 \Rightarrow 16 + 4\lambda + 1 - \lambda + 1 - \lambda = 0 \Rightarrow 2\lambda + 18 = 0 \Rightarrow \lambda = -9$.Substituting $\lambda = -9$ into the family equation: $(4 - 9)x + (-1 - 9)y + (1 - (-9))z - (10 + 4(-9)) = 0 \Rightarrow -5x - 10y + 10z + 26 = 0$,or $5x + 10y - 10z - 26 = 0$.The distance $\alpha$ of the point $(2, 3, -4)$ from this plane is $\alpha = \frac{|5(2) + 10(3) - 10(-4) - 26|}{\sqrt{5^2 + 10^2 + (-10)^2}} = \frac{|10 + 30 + 40 - 26|}{\sqrt{25 + 100 + 100}} = \frac{54}{\sqrt{225}} = \frac{54}{15} = \frac{18}{5}$.Thus,$35\alpha = 35 	imes \frac{18}{5} = 7 	imes 18 = 126$.

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

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