Finding the equation of the new plane after the original has been rotated by an angle 
Find the equation of the plane obtained after rotating the plane $x+y+z=1$ by $90^{\circ}$ about its line of intersection with the plane $x-2y+3z=0$.

Since I had to choose one of the four given choices for this question, I just picked some random points on the line of intersection and found the equation of plane which satisfied them. However, I would like to know how to formally solve this question. Apparently, I must get two planes.
I got the equation of line of intersection as $$\frac{x-\frac23}{5}=\frac{y-\frac13}{-2}=\frac{z}{-3}$$
Now, I need help. The normal of the first plane is obviously perpendicular to this line. After rotation through $90^{\circ}$, will it be parallel to the line? I'm not able to imagine the geometry.
 A: 
Basically there won't really be one possible answer for this question, you can have to two planes that can be rotated by $90$ , the other not shown in the picture. To approach at this I think you will have to take these condition. Since I have had such a question but don't remember exact algorithm now.
1) The rotated plane contain the line. And thus any point on the line as well.
2) The plane is normal to the plane from which is it rotated. (This is the condition which will give two roots , hence two planes).
3)You can also try family of planes. You simply require a plane passing through the intersection of that plane.
This exact approach as far as i remember was quite messy.
$\color{red}{EDIT:}$ I just realized the two planes either by rotating $90$ clockwise or anticlockwise are actually equivalent, there will be only single plane as an answer. Here's why. 
Pictorially, the two "distinct" should be $180$ from each other, but that's just equivalent to $0$ in case of planes. It's only a matter of orientation, and since these are nor vectors having defined normal vector. Both of them are equivalent. 
2) Here's how to approach it algebraically. 
Any plane through intersection of two planes is $(x+y+z-1)+\lambda(x-2y+3z)=0$
It is normal to the plane $(x+y+z-1)=0$ , which gives $\lambda=\frac{-3}{2}$ , notice how there is only single value of $\lambda$.
Hence the required plane is , $x-8y+7z+2=0$
