= 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 $d/\|n\|$. The displacement within the <plane> is perpendicular to $n$.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2018-plane-intersection.png]
{title=A unit <normal vector> and signed plane distance, followed by the <intersection of two planes> as a <straight line>}
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> are
$$
\boxed{x=2,\quad z=2y-2;\qquad r=\begin{pmatrix}2\\0\\-2\end{pmatrix}+t\begin{pmatrix}0\\1\\2\end{pmatrix},\quad t\in\mathbb R.}
$$
There is no valid division by a zero $x$ component in a symmetric ratio form; retaining $x=2$ explicitly handles that component.
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