This is a problem on my homework (so please do not provide more than hints as I definitely don't want you to do it for me). I'm just stuck and was hoping someone might point out my mistake or suggest I do something different.
The problem says I need to find the line of intersection of the two planes as follows: \begin{align} P_1:3z-\left(3x+y\right)=-14,\:\:P_2:2y-4x+4z=-12. \end{align}
I first distributed the negative through on $P_1$ to get \begin{align} 3z-3x-y=-14,\tag{1} \end{align} but now my textbook says that the normal vector to this plane $P_1$ and $P_2$ should be the numbers in front of the $x,y,z$, which would give me the two vectors \begin{align} \vec{v}_1=\left\langle 3,-3,1 \right\rangle,\:\vec{v}_2=\left\langle 2,-4,4 \right\rangle,\tag{2} \end{align} but that is not the case when I graph them (unless somehow I'm graphing them wrong, but I don't see how that's possible...). And when I take the cross product of these two vectors (which I've also checked) I get \begin{align} \vec{v}_1\times \vec{v}_2=\left\langle -16,-14,-6 \right\rangle,\tag{3} \end{align} and this I know to be right, but it isn't at all a scalar multiple of the line of intersection somehow.
Hint: The normal vector to the plane $ax + by + cz = d$ is $\langle a, b, c\rangle$. Note that $3$, the first component of $v_1$, is not the coefficient of $x$ but rather the coefficient of $z$.