How do you find the vector equation of a plane $P=Ax+By+Cz+D$ that makes an angle of $6^{\circ}$ with the $xy$-plane ($z=0$) and has the origin and the position vector $(-3,~9, -1)$ on its plane?
My solution so far: Let $n_1$ be the normal vector of $z=0$ and $n_2$ be the normal vector of $P$. I tried to use the equation: $$\cos\theta = \frac{n_1.n_2}{|n_1||n_2|},$$ I end up getting: $$\cos(6) = \frac{C}{\sqrt{A^2+B^2+C^2}},$$
and I also used the the equation: $$(r-a).n_2=0$$
to get:$$-3A+9B-C=0$$
Now I don't know where to go from here. Am I on the right track?