Let $\overline{a}$  and $\overline{b}$ be two vectors such that $|2\overline{a}+3\overline{b}|=|3\overline{a}+\overline{b}|$  and the angle between  $\overline{a}$ and $\overline{b}$  is ${60}^{\circ }$. If $\frac{1}{8}\overline{a}$  is a unit vector, then $\left|\overline{b}\right|$  is equal to

$|2\overline{a}+3\overline{b}|=|3\overline{a}+\overline{b}|$

S.O.B.S

$4{\overline{a}}^2+9{\overline{b}}^2+\mid 2\overline{a}\cdot \overline{b}=9{\overline{a}}^2+{\overline{b}}^2+6\overline{a}\cdot \overline{b}$

$5{\overline{a}}^2-6\overline{a}\cdot \overline{b}=8|\overline{b}{|}^2$

$5{\overline{a}}^2-6\cdot 8\cdot \left|\overline{b}\right|\cos {60}^{\circ }=8|\overline{b}{|}^2\because \frac{1}{8}|\overline{a}|=1$

$|\overline{b}{|}^2+3|\overline{b}|-40=0$

$\left(\right|\overline{b}|+8)\left(\right|\overline{b}|-5)=0$

$\therefore \left|\overline{b}\right|=5$