Geometry Question - Trihedral angles, planar geometry, spherical geometry 
Two rays, $OX$ and $OY$, are drawn in the horizontal plane $\pi$, and the third ray, $OZ$, is drawn in space so that the rays $OX$, $OY$, and $OZ$ form a trihedral angle $OXYZ$. The planar angles of this trihedral angle are $a = \angle ZOY$, $b = \angle ZOX$, and $c = \angle YOX$. Find the angle between the ray $OZ$ and the plane $\pi$.

I was not sure how to calculate this.

For simplicity, choose $X$, $Y$, $Z$ along their rays such that $|OX|=|OY|=|OZ|=1$. Let $P$ be the foot of the perpendicular dropped from $Z$ into the plane of $\triangle XOY$; then, the angle you seek ---call it "$\theta$"--- is given by
$$\sin\theta=\frac{|PZ|}{|OZ|} = |PZ|$$
We just need a useful expression for $|PZ|$.
Let $Q$ be the foot of the perpendicular dropped from $Z$ onto line $OX$; then, in $\triangle ZOQ$, we have
$$|QZ| = |OZ| \sin \angle ZOX = \sin b$$
Now, $ZQ \perp OX$ by construction; since $ZP \perp \triangle XOY$, we also have $PQ \perp OX$. Thus, in (right) $\triangle ZPQ$, the planar angle at $Q$ matches the dihedral angle along $OX$ (call it "$x$"), and we have
$$\sin\theta = |PZ| = |QZ| \sin \angle ZQP = |QZ| \sin x = \sin b \; \sin x$$
(Had we dropped our perpendicular onto line $OY$, we'd have $\sin\theta = \sin a \; \sin y$. The spherical Law of Sines ---$\frac{\sin a}{\sin x}=\frac{\sin b}{\sin y}=\frac{\sin c}{\sin z}$--- assures us that we have equivalent expressions for $\sin \theta$.)
All that remains is to find $\sin x$ in terms of $a$, $b$, $c$, which is a simple matter of applying the appropriate Spherical Law of Cosines to get
$$\sin\theta = \frac{\sqrt{1+2\cos a \cos b \cos c - \cos^2 a - \cos^2 b - \cos^2 c}}{\sin a}$$

Alternatively (but equivalently), if you know the formula for the volume of a parallelepiped determined by (unit) vectors $\overrightarrow{OX}$, $\overrightarrow{OY}$, $\overrightarrow{OZ}$,
$$\begin{align}
V &= |OX|\;|OY|\;|OZ|\;\sqrt{1+2\cos a \cos b \cos c - \cos^2 a - \cos^2 b - \cos^2 c} \\
&= \sqrt{1+2\cos a \cos b \cos c - \cos^2 a - \cos^2 b - \cos^2 c}
\end{align}$$
then note that that volume is also given by "base-times-height", the product of the area of the parallelogram determined by $OX$ and $OY$, and the length of altitude $PZ$:
$$V = 2|\triangle XOY| \cdot |PZ| = |OX|\;|OY|\;\sin a \cdot h = h \sin a = \sin\theta \; \sin a$$
