= Solution
Provided $p,q$ are <linearly independent>, their linear combinations fill a two-dimensional direction space. Adding $r_0$ translates it to the plane through $r_0$. The <unit normal> and constant are
$$
\boxed{n=\frac{p\times q}{|p\times q|},\qquad c_0=n\cdot r_0.}
$$
Then $n\cdot(r_0+t_1p+t_2q)=c_0$ for every pair of parameters. Conversely a point satisfying $n\cdot r=c_0$ differs from $r_0$ by a vector in $n^\perp=\operatorname{span}\{p,q\}$, proving that the two descriptions give exactly the same plane.
For the three points, choose $r_0=(3,1,-2)$, $p=(-4,1,6)$ and $q=(-1,-2,3)$. Their <cross product> is $(15,6,9)=3(5,2,3)$, which is nonzero. The plane therefore has equation
$$
\boxed{5x+2y+3z=11.}
$$
Each supplied point satisfies it. Its <unit normal> is $(5,2,3)/\sqrt{38}$, up to an overall sign, and $c_0=11/\sqrt{38}$.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2017-plane-and-sphere-section.png]
{title=An affine plane from two direction vectors and a spherical section}
{description=Original geometric illustrations: the supplied three points determine the plane, while a separate schematic shows a generic sphere cut at perpendicular distance d from its centre.}
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