Solution:

$\sqrt{x^{2}+y^{2}+z^{2}}$

Place $A$, $B$, $C$, and $D$ at $(0,0,0)$, $(b, 0,0)$, $(0, c, 0)$, and $(0,0, d)$ in Cartesian coordinate space, with $b$, $c$, and $d$ positive. Then the plane through $B$, $C$, and $D$ is given by the equation $\frac{x}{b}+\frac{y}{c}+\frac{z}{d}=1$. The distance from the origin to this plane is then

$$\frac{1}{\sqrt{\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}}}=\frac{b c d}{\sqrt{b^{2} c^{2}+c^{2} d^{2}+d^{2} b^{2}}}=\frac{b c d}{2 \sqrt{x^{2}+y^{2}+z^{2}}}$$

Then if the area of $BCD$ is $K$, the volume of the tetrahedron is
$$\frac{b c d}{6}=\frac{b c d K}{6 \sqrt{x^{2}+y^{2}+z^{2}}}$$
implying $K=\sqrt{x^{2}+y^{2}+z^{2}}$.

Alternative Solution: The area of $BCD$ is also half the length of the cross product of the vectors $\overrightarrow{BC}=(0,-c, d)$ and $\overrightarrow{BD}=(-b, 0, d)$. This cross product is $(-c d,-d b,-b c)=-2(y, z, x)$, which has length $2 \sqrt{x^{2}+y^{2}+z^{2}}$. Thus the area of $BCD$ is $\sqrt{x^{2}+y^{2}+z^{2}}$.