Calculating DC power usage

You are using \$I_{LINE}\$ in both.

There is no dot, so the current that flows in phase must flow in transmission line. So in a star or wye, \$I_{LINE}\ =\ I_{PHASE}\$ (and \$V_{LINE}\ =\ √ (3)\ V_{PHASE}\$). You show this in your answer.

In a delta, \$I_{LINE}\ =\ √ (3)\ I_{PHASE}\$ (and \$V_{LINE}\ =\ V_{PHASE}\$). A component of two phase currents make up the line current. There is a dot.

$$P_T = √ (3)\ V_{LINE}\ I_{LINE}\ cos\ θ $$ $$P_T = 3\ V_{PHASE}\ I_{PHASE}\ cos\ θ $$

So in your first answer, \$V_{L-L}\$ which is line voltage \$V_{LINE}\ =\ 415V\$, which means a \$V_{PHASE}\ =\ 415V / √ (3)\ =\ 239.6V\$. $$ I_{PHASE}\ =\ V_{PHASE} / Z\ = 239.6V / 10Ω = 24.0A$$ $$ P\ = I_{PHASE}^2\ R = (24.0A)^2\ x\ 8Ω\ =\ 4.59kW$$ $$ P_T\ = 3\ P = 3\ x\ 4.59kW\ =\ 13.8kW$$

1. It depend on contract, usualy only real power is charged, the instrumen shall count real, reactive, apparent.
2. 3 current transformers and 3 volatages is the exact measuring, but you can use one current and one voltage with asumption that network and motor are balanced (that would be rathere an estimation than measuring).
3. Any power meter for rated voltage for your country is ok, let's say 400VAC nominal.

Industrial power meters, there are a lot of them. I can just give you one example: https://www.youtube.com/watch?v=ggDtJWUbkJ4