Interesting puzzle, and more amenable to solving than it first seems.

Label the hands **H**,** M** and ** S**.
- Every 2 minutes
**S** returns to initial position and handedness.
- Every 2 hours
**M** returns to initial position and handedness.
- Every day
**H** returns to initial position and handedness.

So we can't tell how many days have elapsed from the start.

But time of day is all that is asked, so the puzzle should have a solution.

- Initially, and after any even number of minutes
**S** turns CW (and vv.)
- Initially, and after any even number of hours,
**M** turns CW (and vv.)
- Before noon,
**H** turns CW; after noon **H** turns CCW.

So there are two cases for the time = *hh*:*mm*:*ss*.

- Before noon.

**H** turns CW. *hh*=08 (even). **M** turns CW => *mm*=32 (even). **S** turns CW => *ss*=26.

**Time = 08:32:26**

- After noon.

**H** turns CCW. *h*=15 (odd). **M** turns CCW => *mm*=27 (odd). **S** turns CCW => *ss*=34.

**Time = 15:27:34**

**Which is it - all hands moving CW (08:32:26) or all moving CCW (15:27:34)?**

**The only clues I can find are the precise positions of ***H* and *M*.

**If ***M* and *S* are turning CW, *M* would be closer to 6 than to 7 b/c *S* hasn't got to 6 yet.

If *M* and *S* are CCW, *M* would be closer to 6 than to 7 b/c *S* has passed 6.

Either way, *M* should be slightly closer to 6 than to 7.

*M* appears to be slightly closer to 7. Contradiction with either case.

**If ***H* and *M* are turning CW, *H* would be closer to 9 than to 8 b/c *M* has passed 6.

If *H* and *M* are CCW, *H* would be closer to 9 than to 8 b/c *M* hasn't got to 6 yet.

Either way, *H* should be slightly closer to 9 than to 8.

*H* appears to be slightly closer to 9. Consistent with either case.

I'm out of clues: the time is ambiguous.