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(D) The equation of any plane passing through the line of intersection of the planes $P_1: x + y + z - 6 = 0$ and $P_2: 2x + 3y + z + 5 = 0$ is given

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$\frac{11}{\sqrt{5}}$

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(D) The equation of any plane passing through the line of intersection of the planes $P_1: x + y + z - 6 = 0$ and $P_2: 2x + 3y + z + 5 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(x + y + z - 6) + \lambda(2x + 3y + z + 5) = 0$$x(1 + 2\lambda) + y(1 + 3\lambda) + z(1 + \lambda) + (-6 + 5\lambda) = 0$.Since the plane $P$ is perpendicular to the $xy$-plane (whose normal vector is $\vec{k} = (0, 0, 1)$),the normal vector of plane $P$,which is $\vec{n} = (1 + 2\lambda, 1 + 3\lambda, 1 + \lambda)$,must be perpendicular to $\vec{k}$.Therefore,the dot product $\vec{n} \cdot \vec{k} = 0$,which implies $1 + \lambda = 0$,so $\lambda = -1$.Substituting $\lambda = -1$ into the equation of the plane:$x(1 - 2) + y(1 - 3) + z(1 - 1) + (-6 - 5) = 0$$-x - 2y - 11 = 0$,or $x + 2y + 11 = 0$.The distance of the point $(0, 0, 256)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.$d = \frac{|1(0) + 2(0) + 0(256) + 11|}{\sqrt{1^2 + 2^2 + 0^2}} = \frac{11}{\sqrt{1 + 4}} = \frac{11}{\sqrt{5}}$.

Let $P$ be the plane,which contains the line of intersection of the planes $x + y + z - 6 = 0$ and $2x + 3y + z + 5 = 0$ and it is perpendicular to the $xy$-plane. Then the distance of the point $(0, 0, 256)$ from $P$ is equal to

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