If →a, →b, and →c are unit vectors such that →a + 2→b + 2→c = →0, then |→a × →c| is equal to:

The correct option is B
√154
→a + 2→b + 2→c = 0
→a + 2→c = -2→b
Squaring both sides,
(→a + 2→c).(→a + 2→c) = (-2→b).(-2→b)
→a.→a + 4→c.→c + 4→a.→c = 4→b.→b
Since →a, →b, and →c are unit vectors,
1 + 4 + 4→a.→c = 4
5 + 4→a.→c = 4
4→a.→c = -1
→a.→c = -1/4
We know that
|→a × →c|² + (→a.→c)² = |→a|²|→c|²
Since →a and →c are unit vectors,
|→a × →c|² + (-1/4)² = 1² × 1²
|→a × →c|² + 1/16 = 1
|→a × →c|² = 1 - 1/16 = 15/16
|→a × →c| = √(15/16) = √15/4