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(C) The equation of any plane passing through the line of intersection of the planes $P_1: 2x + y - 5z = 0$ and $P_2: 3x - y + 4z - 7 = 0$ is given by

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$(1, 0, 2)$

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(C) The equation of any plane passing through the line of intersection of the planes $P_1: 2x + y - 5z = 0$ and $P_2: 3x - y + 4z - 7 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(2x + y - 5z) + \lambda(3x - y + 4z - 7) = 0$$(2 + 3\lambda)x + (1 - \lambda)y + (-5 + 4\lambda)z - 7\lambda = 0$  --- (Equation $1$)Since the plane is rotated by $\frac{\pi}{2}$ from the original plane $2x + y - 5z = 0$,the normal vectors of these two planes must be perpendicular.The normal vector of the original plane is $\vec{n_1} = (2, 1, -5)$.The normal vector of the new plane is $\vec{n_2} = (2 + 3\lambda, 1 - \lambda, -5 + 4\lambda)$.Since $\vec{n_1} \cdot \vec{n_2} = 0$:$2(2 + 3\lambda) + 1(1 - \lambda) - 5(-5 + 4\lambda) = 0$$4 + 6\lambda + 1 - \lambda + 25 - 20\lambda = 0$$30 - 15\lambda = 0 \Rightarrow \lambda = 2$.Substituting $\lambda = 2$ into Equation $1$:$(2 + 3(2))x + (1 - 2)y + (-5 + 4(2))z - 7(2) = 0$$8x - y + 3z - 14 = 0$.Testing the options,for point $(-1, 0, 2)$ (not in list,checking provided options):For option $D$ $(-1, 0, -2)$: $8(-1) - 0 + 3(-2) - 14 = -8 - 6 - 14 = -28 
eq 0$.Wait,let's re-check the calculation: $8x - y + 3z = 14$. If $x=-1, y=0, z=2$,$8(-1) - 0 + 3(2) = -8 + 6 = -2 
eq 14$. Re-evaluating: The point $(-1, 0, 2)$ satisfies $8(-1) - 0 + 3(2) = -2$ (incorrect). Let's re-check the plane equation: $8x - y + 3z = 14$. Checking point $(-1, 0, 2)$ gives $-8+6 = -2$. Checking $(1, 0, 2)$ gives $8+6=14$. Thus,the point is $(1, 0, 2)$.

If the plane $2x + y - 5z = 0$ is rotated about its line of intersection with the plane $3x - y + 4z - 7 = 0$ by an angle of $\frac{\pi}{2}$,then the plane after the rotation passes through the point

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