= Solution
The center of the <sphere-plane intersection> is the perpendicular projection of the sphere's center onto the plane:
$$
\boxed{\mathbf c_\circ=\mathbf c+\frac{20-\mathbf n\cdot\mathbf c}{|\mathbf n|^2}\mathbf n=(1,2,3)+(1,2,2)=(2,4,5).}
$$
The distance from $\mathbf c$ to the plane is $3<4$, so the intersection is a genuine circle. A <right triangle> with hypotenuse equal to the sphere radius gives
$$
\boxed{\rho=\sqrt{4^2-3^2}=\sqrt7.}
$$
Its plane is the given plane; displacements within it are perpendicular to $\mathbf n$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2-sphere-plane-section.png]
{title=Meridional section of the sphere and its intersection circle}
{description=Original section through the sphere center and the plane normal. The horizontal chord is a diameter of the intersection circle; the right triangle gives $4^2=3^2+\rho^2$. Labels give the centers' original three-dimensional coordinates.}
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