The angles which the vector A=$3\hat{\mathbf{i}}+6\hat{\mathbf{j}}+2\hat{\mathbf{k}}$ makes with the coordinate axes are

If $A=A_{x}i+A_{y}j+A_{z}k$

Then, the angles which it makes with coordinate axes are given by:

$\cos \theta =\frac{\text{ magnitude of component along concerned direction }}{magnitude of A}$

where A is the magnitude of $\overrightarrow{A}=\sqrt{3^2+6^2+2^2}=7$

Therefore, angle with x-axis $\alpha ={\cos }^{-1}\frac{3}{7}$

angle with y-axis $\beta ={\cos }^{-1}\frac{6}{7}$

angle with z-axis $\gamma ={\cos }^{-1}\frac{2}{7}$