Answer:
Option B
Explanation:
Concept involved if a,b,c are any three vectors
then$ |a+b+c|^{2} \geq 0$
$\Rightarrow$ $|a|^{2}+|b|^{2}+|c|^{2}+2(a.b+b.c+c.a) \geq 0$
$\therefore$ $a.b+b.c+c.a \geq \frac{-1}{2} (|a|^{2}+|b|^{2}+|c|^{2})$
Sol.Given $|a-b|^{2}+|b-c|^{2}+|c-a|^{2}=9$
$\Rightarrow$ $|a|^{2}+|b|^{2}-2a.b+|b|^{2}+|c|^{2}-2b.c+|c|^{2}+|a|^{2}-2c.a=9$
$\Rightarrow$ $6-2(a.b+b.c+c.a)=9$ $( \because |a|=|b|=|c|=1)$
$\Rightarrow$ a.b+b.c+c.a=-$\frac{3}{2}$....(i)
Also, a.b+b.c+c.a $\geq \frac{-1}{2} (|a|^{2}+|b|^{2}+|c|^{2}) \geq -\frac{3}{2}$....(ii)
From eq.(i) and (ii) , we get
|a+b+c|=0
as a.b+b.c+c.a is minimum when |a+b+c|=0
$\Rightarrow$ a+b+c=0
$\therefore$ |2a+5b+5c|=|2a+5(b+c)|=|2a-5b|=3
