Angle between a vector and cross product of two vectors 
The angle between the vectors $\overrightarrow {a}$ and $\overrightarrow {b}$ is $\pi/3$, $\overrightarrow {b}$ and $\overrightarrow {c}$ is $\pi/4$, $\overrightarrow {c}$ and $\overrightarrow {a}$ is $\pi/6$.
  Find the angle between $\overrightarrow {a}$ and $\overrightarrow {b}\times \overrightarrow {c}$ 

I tried to calculate the angle by drawing the diagram and geometry but it didn't work out for me. 
The brute Force method of fixing two of the vectors as convenient position vectors also got messy.
How can we evaluate this? I would prefer a general approach, but if there isn't any, a brute Force approach would also be fine.

I just got another solution.
Consider $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ to be unit vectors.
$$\left| \overrightarrow {a}\cdot \left( \overrightarrow {b}\times \overrightarrow {c}\right) \right| ^{2}=\begin{vmatrix} \overrightarrow {a}\cdot \overrightarrow {a} & \overrightarrow {a}\cdot \overrightarrow {b} & \overrightarrow {a}\cdot \overrightarrow {c} \\  \overrightarrow {b}\cdot \overrightarrow {a} & \overrightarrow {b}\cdot \overrightarrow {b} & \overrightarrow {b} \cdot \overrightarrow {c} \\ \overrightarrow {c}\cdot \overrightarrow {a} & \overrightarrow {c}\cdot \overrightarrow {b} &  \overrightarrow {c} \cdot \overrightarrow{a} \end{vmatrix}$$ (product of two determinants)
$$=\begin{vmatrix} 1 & \dfrac {1}{2} & \dfrac {\sqrt {3}}{2} \\ \dfrac {1}{2} & 1 & \dfrac {1}{\sqrt {2}} \\ \dfrac {\sqrt {3}}{2} & \dfrac {1}{\sqrt {2}} & 1 \end{vmatrix}$$
$$=\frac{\sqrt3}{2\sqrt{2}}-\frac{1}{2}$$
$\implies \cos^2(\theta) \times \frac{1}{2} = \frac{\sqrt3}{2\sqrt{2}}-\frac{1}{2}$
$\implies \theta = \cos^{-1} \left(\sqrt{\sqrt{\frac{3}{2}}-1} \right)$ is the required angle.
