For example, given the two intersecting spheres $$S_0: (x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = r_0^2$$ $$S_1: (x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 = r_1^2$$
How would you find the equation of the plane that contains the points in the intersection of spheres $S_0$ and $S_1$?
Rewrite your equations like this for $i\in\{0,1\}$:
$$(x^2+y^2+z^2)+(-2x_i)x+(-2y_i)y+(-2z_i)z =r_i^2-x_i^2-y_i^2-z_i^2$$
Now subtract the two equations from one another, i.e. subtract left side from left side and right side from right side. The $(x^2+y^2+z^2)$ terms will cancel one another and disappear, so you will end up with a linear equation in your coordinates, which is the equation of a plane.
Why does that work? Points which satisfy both the equations must necessarily also satisfy their difference. That's because if the equation is satisfied, both left and right hand side are equal so you are subtracting equal values from equal values. Which means that the points on the circle of intersection will satisfy the difference as well. On the other hand, the difference has the form of a plane, so it must be the plane containing that circle of intersection.
Note that you might have situations where the spheres don't really intersect because they are too far apart. The difference equation would still find a plane between them. The plane that you get from this is called the radical plane of the two spheres. You might also have two concentric spheres, in which case the linear terms would cancel as well, and you'd no longer get the equation of a plane. The radical plane is not defined for concentric spheres.