We have a sphere $(x-1)^2 + (y-1)^2 + (z-1)^2 = 1$ and a point $A = (1;1;-1)$. Find all equations of planes which contain the line $OA$ and intersections with the sphere are circles of radius $\sqrt{3}/3$. Find the centres of the circles.
I don't need the complete solution but just hints in which direction I should try.
Note that a circle on the sphere has radius $\frac{\sqrt{3}}3$ if and only if the center of that circle has a distance of $$\sqrt{1-\left(\frac{\sqrt{3}}3\right)^2}=\sqrt{\frac23}$$ from the center of the given sphere. Hence, all such circles' centers lie on the sphere $$(x-1)^2+(y-1)^2+(z-1)^2=\frac23.$$ Which such circles can be contained in a plane that contains $OA$? Well, the vector from the origin to the circle's center will have to be orthogonal to the circle's normal vector (that is, the normal vector to the smaller sphere at the point chosen for the circle's center), and the vector from $A$ to the circle's center must also be orthogonal to that normal vector , so the possibilities are narrowed greatly. Once you've determined that, you've got the three points that determine the plane in each case: $A$, the origin, and the center of the appropriate circle
Put another way: which tangent planes to the smaller sphere contain both $A$ and the origin?