Find the angle between two planes using their normal vectors 
The angle between two intersecting planes is defined to be the angle between
  their normal vectors. Find the angle between the planes $x – 2y + z = 0$ and $2x + 3y – 2z = 0$. Find the parametric equations of the line of intersection of the two planes above.

For the first plane I said $\overrightarrow n_0 =\langle 1, -2, 1 \rangle$ and for the second plane $\overrightarrow n_1 = \langle 2, 3, -2 \rangle$
Then using $\cos(\theta)= \frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}$ so $\theta = \cos^{-1}\left(\frac{\mathbf A \bullet \mathbf B}{|\mathbf A||\mathbf B|}\right)$
$\mathbf A \bullet \mathbf B = -6$
$|\mathbf A||\mathbf B|= \sqrt{102}$
$\theta = \cos^{-1}\left(\frac{-6}{\sqrt{102}}\right)$
Assuming this is correct so far, how do I find the parametric equations from here?

You can remember that $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$, so that $\vec n_0\times\vec n_1$ gives the direction of the intersection line. In addition, you know that both planes pass through $(0,0,0)$ and so does the intersection line.
