Q.Find the angle between the line $\vec r=(2\hat i-\hat j+\hat k)+t(\hat i+2\hat j-2\hat k)$ and the plane $\vec r\cdot(6\hat i+3\hat j+2\hat k)=8$.

Four closely related "how far / how tilted" computations, all built from a plane's normal $\vec n$ or a line's direction $\vec b$.

Angle between two planes = angle between their normals: $\theta=\cos^{-1}\!\left(\dfrac{|\vec n_1\cdot\vec n_2|}{|\vec n_1||\vec n_2|}\right)$; perpendicular iff $\vec n_1\cdot\vec n_2=0$, parallel iff $\vec n_1=\lambda\vec n_2$.

Angle between a line and a plane = complement of the angle between the line's direction $\vec b$ and the plane's normal $\vec n$ (since a line lying flat in the plane is perpendicular to the normal, and vice versa): $\theta=\sin^{-1}\!\left(\left|\dfrac{\vec b\cdot\vec n}{|\vec b||\vec n|}\right|\right)$; the line is perpendicular to the plane iff $\vec b\parallel\vec n$, and parallel to the plane iff $\vec b\cdot\vec n=0$.

Distance from a point $\vec u$ to a plane $\vec r\cdot\vec n=p$: $\delta=\dfrac{|\vec u\cdot\vec n-p|}{|\vec n|}$ (Cartesian: $\delta=\dfrac{|ax_1+by_1+cz_1-p|}{\sqrt{a^2+b^2+c^2}}$); taking $\vec u=\vec 0$ gives the distance from the origin, $\delta=\dfrac{|d|}{\sqrt{a^2+b^2+c^2}}$ for $ax+by+cz+d=0$. The foot of that perpendicular is $\vec u+\dfrac{p-\vec u\cdot\vec n}{|\vec n|^2}\vec n$.

Distance between two PARALLEL planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ (identical normal direction ratios): $\delta=\dfrac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$ — always rescale one equation first if the normals are only proportional, not identical.

Meeting point of a line $\vec r=\vec a+t\vec b$ and a plane $\vec r\cdot\vec n=p$ (when $\vec b\cdot\vec n\ne0$, i.e. not parallel): substitute the line into the plane's equation, solve the resulting LINEAR equation in $t$ for $t_1=\dfrac{p-\vec a\cdot\vec n}{\vec b\cdot\vec n}$, then the meeting point is $\vec a+t_1\vec b$.