Solution:

(i) AC⊥BD, so BD is perpendicular to the A_{1}AC plane.
If BD is perpendicular to the A_{1}AC plane, then BD is perpendicular to any segment of the plane, and BD⊥ A_{1}C.

(ii) BD is a segment on the line of intersection of planes A_{1}BD and BC_{1}D. We also know that both
A_{1}E and C_{1}E are perpendicular to BD, because they are part of the A_{1}AC plane.
This means that the angle between planes A_{1}BD and BC_{1}D is equal to ∠A_{1}EC_{1}.

∠A_{1}EC_{1} = 180^{o} - ∠A_{1}EA - ∠C_{1}EC.

We already know AA_{1} = CC_{1} = √3. In order to determine ∠A_{1}EA and` ∠C_{1}EC, we need to find AE and CE.

AD⊥DC so triangle ADC is right.

AC^{2} = AD^{2} + DC^{2} = 2^{2} + (2√3)^{2} = 16

AC = 4

cos(∠CAD)= AD/AC = 1/2 so ∠CAD = 60^{o}

cos(∠CAD)= AE/AD, so 1/2 = AE/AD. and AE = 1.

CE = AC - AE = 4 - 1 = 3.

tan(∠AEA_{1}) = AA_{1}/AE = (√3)/1 = √3, so ∠AEA_{1} = 60^{o}.

tan(∠CEC_{1}) = CC_{1}/CE = (√3)/3, so ∠CEC_{1} = 30^{o}.

∠A_{1}EC_{1} = 180^{o} - 60^{o} - 30^{o} = 90^{o}.

The angle between planes A_{1}BD and BC_{1}D is 90^{o}.

(iii) We draw a parallel to AD from point B.

AD and BP are parallel and ∠ADC is right, so ∠BPC is right.

Triangles ADC and ABC are similar, so ∠ACD = ∠ACB = 30^{o}.

∠DCB = 2∠ACD = 60^{o}.

In the right triangle BPC, BP = BC sin60^{o} = 2√3((√3)/2) = 3.

Triangle BC_{1}P is right because BP ⊥ plane CC_{1}D_{1}D.

BC_{1}^{2} = CC_{1}^{2} + BC^{2} = (√3)^{2} + (2√3)^{2} = 3 + 12 = 15.

BC_{1} = √15.

cos(∠PBC_{1})= BP/BC_{1} = 3/√15 = √(3/5).

∠PBC_{1} = arccos(√(3/5)).

In conclusion, the angle formed by lines AD and BC_{1} is equal to arccos(√(3/5)).

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