Any help would be appreciated!
Find a parametric equation of a line $L$ that is obtained as the intersection of the planes $P$ and $Q$ with scalar equations $$P: x+2y+3z=4$$ and $$Q: x-3y+z=1$$
Related to this is the question:
Give a parametric equation of a line $M$ contained in $Q$ and orthogonal to $L$.
Note: This is from a Calculus course so any advice involving linear algebra may not be of much help!
On
Even though this is a question from a Calculus course, it boils down to linear algebra, since the whole problem is a system of linear equations. You need to solve the system
$\begin{align*} x+2y+3z &= 4 \\ x-3y+z &= 1 \end{align*}$
When you solve this, which you can do directly using substitution, you find that
$\begin{align*} x &= -\frac{11}{5}z + \frac{14}{5} \\ y &= -\frac{2}{5}z + \frac{3}{5} \end{align*}$
This is a parametrization of the line you are looking for. If you prefer the vector equation, you could write it as
$$\mathbf{r}(t) = \frac{1}{5}(-11t+14, -2t + 3, t).$$
Normal to the plane P is $\vec p=\vec i+ 2\vec j+ 3\vec k$ Normal to plane Q is $\vec q=\vec i-3 \vec j+\vec k$ The direction vector for the line of intersection is $\vec L=\vec p \times \vec q= 11 \vec i+2 \vec j-5 \vec k.$ For a common point in two plane, let $z=t$ $$x+2y=4-3t~ \& x-3y=1-t \implies y=(3-2t)/5, x=(14-11t)/5$$ Let $t=-1$, the common point is $(5,1,-1)$, the Eq. og line is $$\frac{x-5}{11}=\frac{y-1}{2}=\frac{z+1}{-5}=s.$$ The parametric Eq. of line is $$x=11s+5, y=2s+1, z=-5s-1, s \in \Re$$
EDIT $\vec M. \vec L=0$ and also $\vec M. \vec p=0$ $$\implies \vec M= \vec L \times \vec p=2(8\vec i-19 \vec j + 10\vec k).$$ One can again use the point $(5,1,-1)$ to write the equation of M as $$\frac{x-5}{8}=\frac{y-1}{-19}=\frac{z+1}{10}=u, u \in \Re $$