In a three-dimensional coordinate system with origin $O,$ points $A,$ $B,$ and $C$ are on the $x$-, $y$-, and $z$-axes, respectively.   If the areas of triangles $OAB,$ $OAC,$ and $OBC$ are 4, 6, and 12, respectively, compute the area of triangle $ABC.$
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.

[asy]
import three;

size(250);
currentprojection = perspective(6,3,2);

triple A, B, C, O;

A = (1,0,0);
B = (0,2,0);
C = (0,0,3);
O = (0,0,0);

draw(O--(4,0,0));
draw(O--(0,4,0));
draw(O--(0,0,4));
draw(A--B--C--cycle);

label("$A$", A, S);
label("$B$", B, S);
label("$C$", C, NW);
label("$O$", O,  NE);
[/asy]

Then $\frac{ab}{2} = 4,$ $\frac{ac}{2} = 6,$ and $\frac{bc}{2} = 12,$ so
\begin{align*}
ab &= 8, \\
ac &= 12, \\
bc &= 24.
\end{align*}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 $c = \frac{48}{8} = 6.$

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*}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*}so $[ABC] = \boxed{14}.$