The probable problem with this idea is the channel's poor bending stiffness (these are those Unistruts that the electricians are always playing with, right?). If the parallel channels were very stiff, then the channels would remain perfectly flat under their 250#/2 = 125# load at their midspans. Perfectly flat means that all 6 joists deflect the exact same amount and therefore experience the exact same increase in stresses.
Instead, under your kinda flimsy channels, you can imagine the parallel channels each deforming into a rippled shape, with wave peaks and wave troughs located midway between joists. The wave heights will taper off the further you look from the load. As an interesting thought exercise, think about this ripple pattern but with a total of only 4 joists. In particular, look at an outer joist and the channel wave crest beside it: The flimsy channel is pushing up on the end joist, not pulling down! So in addition to the 125# load itself, now the two interior joists also have additional load from the outer joists pushing down! Holy geez! (I lied a little about the wave peaks and wave trough locations--they're very near midway between joists, but they don't quite hit the midpoints.)
Ceilings are typically designed for 5 psf dead load, where 1/2" drywall plus 2x4 framing on 16" centers leaves 2.5 psf of unused capacity. That extra capacity translates to a parallel channels load at each joist of
(1/8)(2.5psf)(16in)(1ft/12in)(13ft)^2 / [(2.5ft/6.5ft)(1/4)(13ft)] = 57#,
where each of the parallel channels can exert half of this load on the joist. By my reckoning, then, you would need to span (250#)/(57#/joist) = 5 joists with near perfect balance in your lag screw load at every joist.
Sounds like you never go up in that space, so I guess you could gobble up some of the attic's live load capacity for this project. Without knowing more about the attic, it's somewhat safe to assume a 10 psf live load capacity up there. That implies a per-joist available strength of
(1/8)(12.5psf)(16in)(1ft/12in)(13ft)^2 / [(2.5ft/6.5ft)(1/4)(13ft)] = 290#.
Under this schematic a single joist is adequate, leaving a live load capacity in the single joist of
12.5psf - (2.5ft/6.5ft)(1/4)(250#)(13ft) / {(1/8)(13ft)^2[(1joist)(16in/joist)](1ft/12in)} = 1.4 psf.
That's equivalent to 12# applied at midspan, so don't do that.
One consideration when designing a structure is whether loads will occur simultaneously. For instance, you wouldn't design a structure to survive a 50 year return interval wind storm striking at the same time as a 50 year return interval seismic event.
If you do gobble up some live load capacity, then just remember so that if a couple of guys are going up there to insulate or something, you can set the hardtop down on the ground or something.
Oh, and the AWC maintains a connection calculator. that will spec your lag screw withdrawal strength for you (use ASD). If you don't know your joist species, then use Spruce Pine Fir (SPF) for your main member. For the side member use steel. Don't mess with any of the factors at the bottom.