If 𝐚, 𝐛, and 𝐜 are unit vectors satisfying |𝐚−𝐛|² + |𝐛−𝐜|² + |𝐜−𝐚|² = 9, then |2𝐚 + 5𝐛 + 5𝐜| is

|𝐚−𝐛|² + |𝐛−𝐜|² + |𝐜−𝐚|² = 9
⇒ 2|𝐚|² + 2|𝐛|² + 2|𝐜|² − 2(𝐚⋅𝐛 + 𝐛⋅𝐜 + 𝐜⋅𝐚) = 9
(Using |𝐚−𝐛|² = |𝐚|² + |𝐛|² − 2𝐚⋅𝐛)
⇒ 6 − 2(𝐚⋅𝐛 + 𝐛⋅𝐜 + 𝐜⋅𝐚) = 9
⇒ 𝐚⋅𝐛 + 𝐛⋅𝐜 + 𝐜⋅𝐚 = −3/2
Now, |𝐚 + 𝐛 + 𝐜|² = |𝐚|² + |𝐛|² + |𝐜|² + 2(𝐚⋅𝐛 + 𝐛⋅𝐜 + 𝐜⋅𝐚)
⇒ |𝐚 + 𝐛 + 𝐜|² = 3 + 2(−3/2) = 0
𝐚 + 𝐛 + 𝐜 = 0
⇒ 𝐚 = −(𝐛 + 𝐜)
⇒ |2𝐚 + 5(𝐛 + 𝐜)| = |2𝐚 − 5𝐚| = |−3𝐚| = 3|𝐚| = 3