Sequence components of load phase voltages

The sequence components are found with $V_{a0} = \frac13(V_a+V_b+V_c)$, $V_{a1} = \frac13(V_a+aV_b+a^2V_c)$, $V_{a2} = \frac13(V_a+a^2V_b+aV_c)$, where $a = 1\angle120^\circ = -0.5 + j0.866$.

Data: $V_a = 176 - j132$ V, $V_b = -128 - j96$ V (the real part is unclear in the scan; $-128$ is the value in the standard textbook version of this problem), $V_c = -160 + j100$ V.

Zero sequence

$$\begin{aligned} V_{a0} &= \tfrac13\left[(176 - 128 - 160) + j(-132 - 96 + 100)\right] \\ &= \tfrac13(-112 - j128) = -37.33 - j42.67 = 56.69\angle{-131.19^\circ}\ \text{V} \end{aligned}$$

Positive sequence

$$\begin{aligned} aV_b &= (1\angle120^\circ)(160\angle{-143.13^\circ}) = 160\angle{-23.13^\circ} = 147.14 - j62.85 \\ a^2V_c &= (1\angle240^\circ)(188.68\angle148.0^\circ) = 188.68\angle28.0^\circ = 166.60 + j88.56 \\ V_{a1} &= \tfrac13\left[(176 + 147.14 + 166.60) + j(-132 - 62.85 + 88.56)\right] \\ &= \tfrac13(489.74 - j106.29) = 163.25 - j35.43 = 167.05\angle{-12.24^\circ}\ \text{V} \end{aligned}$$

Negative sequence

$$\begin{aligned} a^2V_b &= 160\angle96.87^\circ = -19.14 + j158.85 \\ aV_c &= 188.68\angle268.0^\circ = -6.60 - j188.56 \\ V_{a2} &= \tfrac13\left[(176 - 19.14 - 6.60) + j(-132 + 158.85 - 188.56)\right] \\ &= \tfrac13(150.26 - j161.71) = 50.09 - j53.90 = 73.58\angle{-47.10^\circ}\ \text{V} \end{aligned}$$

Check

$V_{a0} + V_{a1} + V_{a2} = (-37.33 + 163.25 + 50.09) + j(-42.67 - 35.43 - 53.90) = 176.0 - j132.0 = V_a$ ✓

Components of phases b and c: $V_{b1} = a^2V_{a1}$, $V_{c1} = aV_{a1}$, $V_{b2} = aV_{a2}$, $V_{c2} = a^2V_{a2}$, $V_{b0} = V_{c0} = V_{a0}$.

Answer: $V_1 = 167.05\angle{-12.24^\circ}$ V, $V_2 = 73.58\angle{-47.10^\circ}$ V, $V_0 = 56.69\angle{-131.19^\circ}$ V (phase a reference).