If |\(\vec{a}\)|=3|\(\vec{b}|,|\vec{b}\)|=2 and angle between \(\vec{a}\) and \(\vec{b}\) is 60^o,then  |\(\vec{a}-\vec{b}\)| is equal to:

2\(\sqrt { 7}\)

$|\vec{a}|=3|\vec{b}|,|\vec{b}|=2$  i.e. $|\vec{a}| =3×2=6$

let θ be angle between $\vec{a}\, \And\, \vec{b}$ then θ = 60^o

$|\vec{a}-\vec{b}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\vec{b}}$

$=\sqrt{36+4-2×\frac{1}{2}×2}=\sqrt{40-12}=\sqrt{28}=2\sqrt{7}$

So option 2 is correct.