Past exam of the mathematical methods I course of the University of Oxford / 2018 / 7 / c / Solution

A plane through $r_0$ with nonzero normal vector $n$ consists of the position vectors $r$ satisfying $(r-r_0)\cdot n=0$. Thus $r\cdot n=d$, with $d=r_0\cdot n$. When $n$ is a unit vector, $d$ is the signed distance from the origin in the direction $n$; for an unnormalized normal vector, that distance is $d/\Vert n\Vert$. The displacement within the plane is perpendicular to $n$.

The cross product of the two normal vectors is $(2,6,-3)\times (5,2,-1)=(0,-13,-26)$, so their intersection of two planes has direction $(0,1,2)$. Eliminating $2y-z$ gives $x=2$, and substitution then gives $z=2y-2$. Thus the Cartesian constraints and parametric equation of a straight line areThere is no valid division by a zero $x$ component in a symmetric ratio form; retaining $x=2$ explicitly handles that component.