Question

Consider the following planes. 3x − 2y + z = 1, 2x + y − 3z = 3

Answer 1

Solve for any of the variables; for instance, $z$:

$z=1-3x+2y$

$z=-\dfrac{3-2x-y}3$

Then

$1-3x+2y=-\dfrac{3-2x-y}3\implies11x-5y=6\implies\begin{cases}y=\frac{11x-6}5\\z=1-3x+\frac{22x-12}5\end{cases}$

Let $x=t$; then the intersection is given by the vector-valued function

$\vec r(t)=\left(t,\dfrac{11t-6}5,1-3t+\dfrac{22t-12}5\right)$

or

$\vec r(t)=\left(t,\dfrac{11t-6}5,\dfrac{7t-7}5\right)$