Could someone check my work for this problem for vectors? Find the angle  between the two planes. (Round your answer to two decimal places. If the planes do not intersect, enter DNE.)
I found this to be 65.28°
(b) Find parametric equations of their line of intersection. (Let $z = t$, then solve for $x(t)$ and $y(t)$ in terms of $t$. If the planes do not intersect, enter DNE.)
$x −  3y + z    =  −5$
$3x  +  2z +  5  =  0$
My try: $3x + 2z + 5 = 0  ⇒  x = \frac{1}{3}(-2z-5)$
Then $3y = x + z + 5$
$3y= \frac{1}{3}(2z-5)+z+5 =  \frac{z+10}{3} \implies y = \frac{z+10}{9}$
Let $z = t$. Parametric equations: $x=-\frac{2t}{3}-\frac{5}{3},\ y=\frac{t}{9}+\frac{10}{9},\ z=t$

$$\cos^{-1} \left( \frac{3+2}{\sqrt{(1^2+3^2+1)(3^2+2^2)}}\right)=\cos^{-1}\left( \frac{5}{\sqrt{143}}\right)\approx 65.28^\circ$$
To verify your solution, we just have to substitute them in:
$$x-3y+z=-\frac{2t}{3}-\frac53-\frac{t}3-\frac{10}3+t=-5$$
$$3x+2z+5=-2t-5+2t+5=0$$
It is fine.
