3D Geometry Question In $3$-dimensional Geometry, if angle made of line segment $OP$ with $X,Y,Z$-axis are in 
$1:2:3$, then what is the angle made by line segment with $Y$-axis?
My Solution:
Let $\alpha,\beta$ and $\gamma$ be the angle made by line with $X,Y,Z$-axis, respectively. Then
$$\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma = 1$$
Now given $\alpha = \theta$ and $\beta = 2\theta$ and $\gamma = 3\theta$,
So $\cos^2 \theta+\cos^2 2\theta+\cos^2 3\theta = 1$,
or 
$$2\cos^2 \theta+2\cos^2 2\theta+2\cos^2 3\theta = 2,$$
or 
$$1+\cos 2\theta+\cos^2 2\theta+1+\cos 6\theta = 2,$$
or
$$1+\cos 2\theta+2\cos^2 2\theta+1+4\cos^3 2\theta-3\cos 2 \theta = 2.$$
Let $\cos 2 \theta = y$, then
$$y+2y^2+4y^3-3y = 0,$$
$$4y^3+2y^2-2y = 0,$$
$$2y(2y^2-y-1) = 0.$$
Maybe I have missed something, but I feel that this method is very tedious.
So could anyone explain me a better method?
Thanks.

say smallest angle is $\theta$ 
therefore $\theta+2\theta +3\theta= 180^\circ$. 
Sum of the angles of a triangle equal $180^\circ$.
therefore $\theta=30^\circ$ (since $6\theta=180^\circ=> \theta=\dfrac{180}{6}=30^\circ$)
therefore angle made with y axis is $60^\circ$.
