# How to find an equation of a plane perpendicular to two other planes and passing through a point

Please, could anybody help me with the next problem. I have two planes:

$$2x-y+5z+3=0 \ (\text{red plane})\\ x+3y-z-7=0 \ (\text{green plane})$$ And I need to find a plane which is perpendicular to these planes and passes through a point $O(0,0,0)$.

At first I thought it can be done via finding the cross product of the normal vectors of these planes $n_1(2,-1,5)×n_2(1,3,-1) = (-14,6,7)$ and then substitute these to the standard plane equation $-14x+6y+7z=0$. But this result is not correct, the brown plane is the result and it is not perpendicular to the red one, though it contains the point $O$, which is the small green ball on the picture: Also I saw these two questions, which are the same as mine:

Find a plane that passes through a point and is perpendicular to 2 planes

Find the equation of a plane

But the accepted answer suggests the same solution I tried myself, and it seems to be wrong.

Any help is greatly appreciated, thank you in advance.

Your calculation of the cross product is incorrect. You should have $n_1\times n_2 = (-14, 7, 7)$. I imagine, once you fix that, you should have the plane you desire as you are using the correct method.