Symmetrical components of unbalanced voltages

Using $a = 1\angle120^\circ$: $V_0 = \frac13(V_a+V_b+V_c)$, $V_1 = \frac13(V_a+aV_b+a^2V_c)$, $V_2 = \frac13(V_a+a^2V_b+aV_c)$.

Rectangular form

$$\begin{aligned} V_a &= 200\angle0^\circ = 200 + j0 \\ V_b &= 200\angle245^\circ = -84.52 - j181.26 \\ V_c &= 200\angle105^\circ = -51.76 + j193.19 \end{aligned}$$

Zero sequence

$$V_0 = \tfrac13(63.71 + j11.92) = 21.24 + j3.97 = 21.61\angle10.60^\circ\ \text{V}$$

Positive sequence

$aV_b = 200\angle365^\circ = 200\angle5^\circ$, $a^2V_c = 200\angle345^\circ = 200\angle{-15^\circ}$:

$$\begin{aligned} V_1 &= \tfrac13(200\angle0^\circ + 200\angle5^\circ + 200\angle{-15^\circ}) \\ &= \tfrac13(592.42 - j34.33) = 197.47 - j11.44 = 197.81\angle{-3.32^\circ}\ \text{V} \end{aligned}$$

Negative sequence

$a^2V_b = 200\angle485^\circ = 200\angle125^\circ$, $aV_c = 200\angle225^\circ$:

$$\begin{aligned} V_2 &= \tfrac13(200\angle0^\circ + 200\angle125^\circ + 200\angle225^\circ) \\ &= \tfrac13(-56.14 + j22.41) = -18.71 + j7.47 = 20.15\angle158.24^\circ\ \text{V} \end{aligned}$$

Check: $V_0 + V_1 + V_2 = 200.0 + j0 = V_a$ ✓

Answer: $V_0 = 21.61\angle10.60^\circ$ V, $V_1 = 197.81\angle{-3.32^\circ}$ V, $V_2 = 20.15\angle158.24^\circ$ V. The positive sequence dominates because the set is nearly balanced (only the angles are unbalanced).