Parametric and Cartesian equations of lines

The line through two points has as direction vector the vector $\overrightarrow{AB}$, obtained by subtracting the coordinates of $A$ from those of $B$. The parametric equations are written starting from the point $A$, adding to each coordinate $t$ times the corresponding component of $\vec{v}$. For the Cartesian equations the parameter is eliminated. The second equation states directly $t=y$, so substituting into the first and the third gives $x=1+2y$ and $z=2-3y$, that is, two planes containing the line. As a check, the point $B=(3,1,-1)$ satisfies both equations, since $3-2-1=0$ and $3-1-2=0$.

The line is the intersection of two planes, and the normal vectors ${\vec{n}}_1=(1,1,-1)$ and ${\vec{n}}_2=(1,-1,2)$ are not parallel, so the system does represent a line. The direction vector is orthogonal to both normal vectors, therefore we take $\vec{v}={\vec{n}}_1\times {\vec{n}}_2=(1,-3,-2)$. To find a point we fix one coordinate: with $x=0$ the system becomes two equations in $y$ and $z$, and adding them gives $z=3$, hence $y=2+z=5$. The point $(0,5,3)$ and the vector $\vec{v}$ give the parametric equations. As a check, the generic point satisfies both equations for every $t$, since $t+(5-3t)-(3-2t)=2$ and $t-(5-3t)+2(3-2t)=1$.

Parallel lines have parallel direction vectors, so a direction vector of $s$ is needed. The line $s$ is given in Cartesian form, therefore its direction vector is the cross product of the two normal vectors, $(1,3,5)$. The parametric equations of $r$ are written with the point $P$ and the same direction vector. For the Cartesian equations we solve the first equation for $t=x-1$ and substitute it into the other two. As a check, the point $P$ satisfies the two equations, since $3-2-1=0$ and $5+1-6=0$, and the normal vectors $(3,-1,0)$ and $(5,0,-1)$ are both orthogonal to $(1,3,5)$.