3. Given in the tetrahedron $S-ABC$, $SC \perp CB$, $SA \perp AB$, $CB \perp AB$, and the angles formed by $SA$, $SC$ with the plane of $\triangle ABC$ are equal. If $AC=6$, and the distance from $S$ to the plane $ABC$ is 4, then the distance between the skew lines $AC$ and $SB$ is ( ).
(A) $\frac{6 \sqrt{13}}{13}$
(B) $\frac{12 \sqrt{13}}{13}$
(C) $\frac{12}{5}$
(D) $\frac{24}{5}$
3. A.

As shown in Figure 3, draw $SD \perp$ plane $ABC$, with the foot of the perpendicular being $D$. Connect $AD$, $CD$, and $BD$, and let $AC$ intersect $BD$ at point $O$. Since $SC \perp CB$, according to the Three Perpendiculars Theorem, we have $CD \perp BC$. Similarly, $AB \perp AD$. Also, since $CB \perp AB$, quadrilateral $ABCD$ is a rectangle. Given that the angles formed by $SA$ and $SC$ with the plane of $\triangle ABC$ are equal, we have $AD = CD$. Therefore, quadrilateral $ABCD$ is a square.
Draw $OE \perp SB$, with the foot of the perpendicular being $E$.
Since $SD \perp AC$ and $AC \perp BD$, we have $AC \perp$ plane $SBD$. Thus, $AC \perp OE$. Therefore, $OE$ is the common perpendicular of $AC$ and $SB$.
Draw $DF \perp SB$, with the foot of the perpendicular being $F$. Then we have
$$
OE = \frac{1}{2} DF = \frac{1}{2} \cdot \frac{SD \cdot BD}{SB} = \frac{1}{2} \times \frac{4 \times 6}{2 \sqrt{13}} = \frac{6 \sqrt{13}}{13}.
$$

Therefore, the distance between the skew lines $AC$ and $SB$ is $\frac{6 \sqrt{13}}{13}$.