Can a vector have direction angles $30^{\circ}$, $45^{\circ}$, $60^{\circ}$?
Let $\alpha = 30^{\circ}$, $\beta = 45^{\circ}$, $\gamma = 60^{\circ}$
direction cosines are: $\;\;$ $\ell = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$
$\therefore$ $\;$ $\ell = \cos 30^{\circ} = \dfrac{\sqrt{3}}{2}$, $\;$ $m = \cos 45^{\circ} = \dfrac{1}{\sqrt{2}}$, $\;$ $n = \cos 60^{\circ} = \dfrac{1}{2}$
Now, $\;$ $\ell^2 + m^2 + n^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 + \left(\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{1}{2}\right)^2 = \dfrac{3}{4} + \dfrac{1}{2} + \dfrac{1}{4} = \dfrac{3}{2}$
$\because$ $\;$ $\ell^2 + m^2 + n^2 \neq 1$, $\;$ $\therefore$ $\;$ the vector cannot have direction angles $30^{\circ}$, $45^{\circ}$ and $60^{\circ}$.