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

The plane $$2x + y - 5z = 0$$ is rotated about its line of intersection with the plane $$3x - y + 4z - 7 = 0$$ by an angle $$\frac{\pi}{2}$$.

First, any plane through the line of intersection of the two given planes can be written as: $$(2x + y - 5z) + \lambda(3x - y + 4z - 7) = 0.$$ Expanding this expression gives $$(2 + 3\lambda)x + (1 - \lambda)y + (-5 + 4\lambda)z - 7\lambda = 0.$$

Next, since the rotated plane is obtained by rotating the original plane by $$\frac{\pi}{2}$$, it must be perpendicular to the original plane $$2x + y - 5z = 0.$$ The normal to the original plane is $$\vec{n_1} = (2, 1, -5)$$, while the normal to the rotated plane is $$\vec{n} = (2 + 3\lambda,\,1 - \lambda,\,-5 + 4\lambda).$$ For perpendicularity we require $$\vec{n} \cdot \vec{n_1} = 0$$, which leads to $$2(2+3\lambda) + 1(1-\lambda) + (-5)(-5 + 4\lambda) = 0,$$ $$4 + 6\lambda + 1 - \lambda + 25 - 20\lambda = 0,$$ and therefore $$30 - 15\lambda = 0 \implies \lambda = 2.$$

Now substituting $$\lambda = 2$$ into the equation $$(2 + 3\lambda)x + (1 - \lambda)y + (-5 + 4\lambda)z - 7\lambda = 0$$ yields $$(2 + 6)x + (1 - 2)y + (-5 + 8)z - 14 = 0,$$ which simplifies to $$8x - y + 3z - 14 = 0.$$

Finally, checking which point satisfies this equation shows that Option A: $$(2, -2, 0)$$ gives $$16 + 2 + 0 - 14 = 4 \neq 0,$$ Option B: $$(-2, 2, 0)$$ gives $$-16 - 2 + 0 - 14 = -32 \neq 0,$$ and Option C: $$(1, 0, 2)$$ gives $$8 - 0 + 6 - 14 = 0$$ $$\checkmark$$. Therefore the correct answer is Option C: $$(1, 0, 2)$$.