jmc

We can think of $TAB$ as the base of the pyramid, and $\overline{CT}$ as the height from apex $C$ to the base, since $\overline{CT}$ is perpendicular to face $ABT$. The area of right triangle $ABT$ is $(12)(12)/2=72$ square units, so the volume of the pyramid is $\frac{1}{3}([ABT])(CT)=\frac{1}{3}(72)(6)=144$ cubic units.

Letting the distance from $T$ to face $ABC$ be $h$, the volume of $TABC$ can also be expressed as $\frac{h}{3}([ABC])$, so $\frac{h}{3}([ABC])=144$, from which we have $$h=\frac{432}{[ABC]}.$$Applying the Pythagorean Theorem to triangles $TAB$, $TAC$, and $TBC$, we have $$\begin{array}{rl}AB& =12\sqrt{2},\\ AC& =BC=\sqrt{12^2+6^2}=\sqrt{6^2(2^2+1^2)}=6\sqrt{5}.\end{array}$$Therefore, $△ABC$ is isosceles. Altitude $\overline{CM}$ of $△ABC$ bisects $\overline{AB}$, so we have $AM=6\sqrt{2}$. Applying the Pythagorean Theorem to $△ACM$ gives us $CM=6\sqrt{3}$, so $$[ABC]=\frac{(AB)(CM)}{2}=36\sqrt{6}.$$Substituting this into our equation for $h$ above, we have $$h=\frac{432}{[ABC]}=\frac{432}{36\sqrt{6}}=\frac{36\cdot 12}{36\sqrt{6}}=\frac{12}{\sqrt{6}}=\boxed{2\sqrt{6}}.$$