5. Given a tetrahedron with three lateral edges that are pairwise perpendicular, and the angles formed by each lateral edge with the base are $\alpha, \beta, \gamma$. Then, $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\sqrt{2}$
5. C. The projection of point W onto the plane of $\triangle ABC$ is the orthocenter of $\triangle ABC$, and $AD \perp \mathcal{Z}$. Since $PD \perp BC$, therefore, in the right triangle $\triangle PAD$, $\angle PAD = \alpha$, and since $D = \frac{bc}{\sqrt{b^2 + c^2}}$, $\operatorname{tg} c = \frac{bc}{\sqrt{a^2 b^2 + a^2 c^2}}$.
Then
$$
\sin^2 \alpha
= \frac{b^2 c^2}{a^2 b^2 + b^2 c^2 + c^2 a^2} \text{.}
$$

Similarly, $\sin^2 \beta =$
$$
\begin{array}{l}
\frac{c^2 a^2}{a^2 b^2 + b^2 c^2 + c^2 a^2}, \\
\sin^2 \gamma = \frac{a^2 b^2}{a^2 b^2 + b^2 c^2 + c^2 a^2} . \\
\therefore \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1 .
\end{array}
$$