Given that the equation of the line is:
$$
\mbox{P:}\quad
\left\{\begin{array}{rcrcrcr}
3x & – & y & + & z & = & 6
\\
x & + & 2y & + & z & = &-3
\end{array}\right.
$$
$$
\mbox{and the plane is A:}\quad
x + 2y + z = 5.
$$
I believe you need to find the vector and use it to find the angle between the vector of the line and the normal vector of the plane.
I tried finding two points for the first equation but couldn't move further from there.
Best Answer
The vectors (3,-1,1) and (1,2,1) are the normals to the two planes that describe the line. Hence, the vector along the line will be the cross product of these.
Similarly, the plane's normal vector is (1,2,1) (which in your example contains the line, so the angle is zero).
But in general, you can fine the angle, $\theta$ between the line and normal to the plane via
$$\cos(\theta) = \frac{\vec{p} . \vec{l}}{(||\vec{p}||) (||\vec{l}||)}$$
Where $\vec{l}$ and $\vec{p}$ are the vectors describing the direction of the line and the normal of the plane respectively.
and then $\frac{\pi}{2}- \theta$ is the angle you're after.