3. The following is a 3-dimensional composite figure:

Step 1 ：Let $A = (a,0,0),$ $B = (0,b,0),$ and $C = (0,0,c).$ Without loss of generality, we can assume that $a,$ $b,$ and $c$ are positive.

Step 2 ：Then $\frac{ab}{2} = 4,$ $\frac{ac}{2} = 6,$ and $\frac{bc}{2} = 12,$ so

Step 3 ：\begin{align*} ab &= 8, \\ ac &= 12, \\ bc &= 24. \end{align*}

Step 4 ：Multiplying all these equations, we get $a^2 b^2 c^2 = 2304,$ so $abc = 48.$ Hence, $a = \frac{48}{24} = 2,$ $b = \frac{48}{12} = 4,$ and $c = \frac{48}{8} = 6.$

Step 5 ：Then \begin{align*} AB &= \sqrt{a^2 + b^2} = 2 \sqrt{5}, \\ AC &= \sqrt{a^2 + c^2} = 2 \sqrt{10}, \\ BC &= \sqrt{b^2 + c^2} = 2 \sqrt{13}. \end{align*}

Step 6 ：By Heron's Formula, \begin{align*} [ABC]^2 &= (\sqrt{5} + \sqrt{10} + \sqrt{13})(-\sqrt{5} + \sqrt{10} + \sqrt{13})(\sqrt{5} - \sqrt{10} + \sqrt{13})(\sqrt{5} + \sqrt{10} - \sqrt{13}) \\ &= ((\sqrt{10} + \sqrt{13})^2 - 5)(5 - (\sqrt{10} - \sqrt{13})^2) \\ &= (2 \sqrt{130} + 18)(2 \sqrt{130} - 18) \\ &= 196, \end{align*}

Step 7 ：so $\boxed{[ABC] = 14}.$