Let ω be a solution of x³ - 1 = 0 with Im(ω) > 0. If a = 2 with b and c satisfying (E), then the value of 3ωa + 1ωb + 3ωc is equal to:?

Since, a, b and c satisfies ⎧⎩
a
b
c
⎨⎪ ⎧⎩
1 9 7
8 2 7
7 3 7
⎨⎪ = ⎧⎩
0
0
0
⎨⎪

So, we get the equations
a + 8b + 7c = 0
9a + 2b + 3c = 0
7a + 7b + 7c = 0 or a + b + c = 0
Since, a = 2, so the equations become
2 + 8b + 7c = 0 (1)
18 + 2b + 3c = 0 (2)
2 + b + c = 0 (3)
Solving the above equations, we get
b = 12, c = -14
3ωa + 1ωb + 3ωc = 3ω2 + 1ω12 + 3ω-14 = 3ω2 + 1(ω3)4 + 3(ω3)-4ω2 = 3ω2 + 1 + 3ω2 = 3ω + 1 + 3ω2
Since 1 + ω + ω2 = 0,
So, 3ωa + 1ωb + 3ωc = 3(-1) + 1 = -2