The planes and are not parallel, so they must intersect along a line that is common to both of them. The vector parametric equation for this line is
I cannot seem to know where to begin this problem.
Hello, r2d2!
$\displaystyle \text{The planes }\:\begin{Bmatrix}3y - 3x - 4z &=& \text{-}18 & [1] \\ 3x - 2y + 3z &=& 14 & [2] \end{Bmatrix}\text{ are not parallel,}$
$\displaystyle \text{so they must intersect along a line that is common to both of them.}$
$\displaystyle \text{The vector parametric equation for this line is }\_\_.$
There are a number of ways to solve this problem.
Here is one of them.
$\displaystyle \begin{array}{cccccccc}\text{Multiply [1] by 3:} & \text{-}12x + 9y - 12z &=& \text{-}54 & [3] \\ \text{Multiply [2] by 4:} & 12x - 8y + 12z &=& 56 & [4]\end{array}$
Add [3] and [4]: .$\displaystyle y \:=\:2$
Substitute into [4]: .$\displaystyle 12x - 8(2) + 12z \:=\:56 \quad\Rightarrow\quad x \:=\:6-z$
. . We have: .$\displaystyle \begin{Bmatrix} x &=& 6-z \\ y &=& 2 \\ z &=& z \end{Bmatrix}$
$\displaystyle \text{On the right, replace }z\text{ with a parameter }t\!: \;\;\begin{Bmatrix}x &=& 6 - t \\ y &=& 2 \\ z &=& t \end{Bmatrix}$