Angle between two planes and between a straight line and a plane

We need normal vectors of both planes. To obtain a perpendicular vector to $\pi_1$, it is necessary to compute the vector product of its two governing vectors:

$$\vec{n}_1=\begin{vmatrix} and & j & k \\ 1 & -2 & 4\\ 3 & 0 & 1 \end{vmatrix} = -2i+12j+6k-j=-2i+11j+6k=(-2,11,6)$$

$$\vec{n}_2=(2,-11,6)$$

and we can already solve:

$$\begin{array}{rl} \alpha =& \arccos \Big( \dfrac{|A_1 A_2+B_1 B_2+C_1 C_2|} {\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}} \Big) \\ =& \arccos\Big( \dfrac{|(-2)\cdot2+11\cdot(-11)+6\cdot6|} {\sqrt{(-2)^2+11^2+6^2}\sqrt{2^2+(-11)^2+6^2}} \Big) \\ =& \arccos \Big( \dfrac{21}{161}\Big)= \arccos(0.1304)= 82.51^\circ \end{array}$$

A governing vector of the straight line is $\vec{v}=(0,1,3)$.

A normal vector to the plane is $\vec{n}= (3, -1, 2)$.

Therefore we can apply the formula:

$$\begin{array}{rl} \alpha=& \arcsin\Big( \dfrac{|v_1 A+v_2 B+v_3 C|} {\sqrt{v_1^2+v_2^2+v_3^2}\sqrt{A^2+B^2+C^2}} \Big) \\ =& \arcsin\Big( \dfrac{|0\cdot3+1\cdot(-1)+3\cdot2|} {\sqrt{0^2+1^2+3^2}\sqrt{3^2+(-1)^2+2^2}} \Big) \\ =& \arcsin\Big( \dfrac{5}{\sqrt{140}} \Big)= \arcsin{0.4226}=25^\circ \end{array}$$