3. In the tetrahedron $D-ABC$, it is known that $AB=$ $BC=2, AB \perp BC, BC \perp CD, DA \perp AB$, $\angle CDA=60^{\circ}$. Then the volume of the tetrahedron $D-ABC$ is $\qquad$
3. $\frac{4}{3}$.

As shown in Figure 2, draw $DE \perp$ plane $ABC$ at point $E$, and connect $EA$, $EC$.

Since $BC \perp CD$, $DA \perp AB$, therefore,
$EC \perp CB$,
$EA \perp AB$.
Thus, quadrilateral $EABC$ is a rectangle.

Given $AB = BC$, we know that rectangle $EABC$ is a square, and $DA = DC$.
Also, $\angle CDA = 60^\circ$, so $\triangle DAC$ is an equilateral triangle.
Thus, $DA = AC$.
Therefore, $\sqrt{EA^2 + ED^2} = \sqrt{EA^2 + EC^2}$
$\Rightarrow ED = EC = 2$.
Hence, the volume of the tetrahedron $D-ABC$ is
$\frac{1}{3}\left(\frac{1}{2} \times 2 \times 2\right) \times 2 = \frac{4}{3}$.