A load that has a lagging power factor is, by convention, said to be receiving reactive power from the source. A load that has a leading power factor is, by convention, said to be delivering reactive power to the source. In a vector representation of AC circuits, inductors are given the value +jX and capacitors are given the value -jX, so that leads to the the use of +jI for lagging reactive currents and -jI for leading reactive currents. Therefore, lagging reactive power is positive and leading reactive power is negative.
Sometimes a leading power factor is given a negative sign, but that is not a good practice, because power = V X I X pf and negative real power is power that is returned to the source by the load as a motor does during regenerative braking or a utility customer does by generating more power than they use with a renewable energy system.
Solution is all but trivial.
A rather good starting point would be doing the following assumptions valid for a resistive load:
1)Each cycle will start with C1 fully discharged and open TRIAC U2.
2)Load resistance is much lower than R3+RV1, hence you will have full half sine mains voltage across TRIAC.
3)DIAC is open circuit untill its breakover voltage VBO (approx 30V) is reached.
4)Now we have to write the transient of C1 being fed with sine voltage via R3+RV1.
5)When vc(t)=VBO TRIAC is fired, capacitor is discharged and your load is being fed.
So KVL to the source, R, C mesh would be
$$V_\text{max}\sin\omega t=v_\text{C}(t)+RC\,\frac{\text{d}\,v_\text{C}(t)}{\text{dt}}$$
the usual first order ODE to be solved in \$v_\text{C}(0)=0\$ boundary (or more generally some initial voltage as per Spehro's comment).
This is known to have solution sum of its general \$\breve{v}_\text{C}(t)=A\,\text{e}^{-t/\tau}\quad\tau=RC\$
and particular one \$\hat{v}_\text{C}(t)=\frac{V_\text{max}}{\sqrt{1+\omega^2\tau^2}}\sin\left(\omega t- \arctan(\omega\, \tau)\,\right)\$
Combining them in the above constraint and applying a little trigo gives
$$v_\text{C}(t)=\frac{V_\text{max}}{\sqrt{1+\omega^2\tau^2}}\sin\left(\omega t- \arctan(\omega\, \tau)\,\right)+\frac{V_\text{max}\,\omega\, \tau}{1+\omega^2\tau^2}\text{e}^{-t/\tau}$$
which equated to DIAC break over would give TRIAC on time.
$$\frac{V_\text{BO}}{V_\text{max}}=\frac{\sin(\omega\, t_\text{on}- \arctan(\omega\, \tau)\,)}{\sqrt{1+\omega^2\tau^2}}+\frac{\omega\,\tau\,\text{e}^{-t_\text{on}/\tau}}{1+\omega^2\tau^2}$$
What we really understand from the above is that's indeed job for a numeric solutor.
Edit: one sign fixed upon @Delfin suggestion.
Best Answer
In a balanced three phase system, the power in each of the three phases behaves in the same way as it does in single-phase systems. When you add together the power in the three phases, the result is constant rather than pulsating. The reactive energy flows back and forth between storage elements in each of the three phases of the load and storage elements in the source or distribution system just as it does in a single phase system.