Let us assume that after the collision, the velocity of the incoming ball changes from $u$ to $u^{\text{'}}$ and the other two balls move in the opposite directions with the same speed $v$, then

$-2\left(N∆t\right)\cos 30°= m{u}^{\text{'}}-mu$

$\Rightarrow -\left(N∆t\right)\sqrt{3}=m{u}^{\text{'}}-mu$

For the ball attached to the string,

$\left(N∆t\right)\sin 30°=mv$

$\Rightarrow N∆t=2mv$

Eliminating $N∆t$, we obtain

$-2\sqrt{3}mv=m{u}^{\text{'}}-mu$

Using the coefficient of restitution equation we get

$1=\frac{v\cos 60°-u^{\text{'}}\cos 30°}{u\cos 30°}$

$\Rightarrow \sqrt{3}u=v-\sqrt{3}{u}^{\text{'}}$

$\Rightarrow 2\sqrt{3}v=6u+6u^{\text{'}}$

$6u+6u^{\text{'}}=u-u^{\text{'}}$

$\Rightarrow u^{\text{'}}=-\frac{5}{7}u$

The total impulse of tension is

$-2T∆t=m{u}^{\text{'}}-mu$

$2T∆t=m\frac{5u}{7}+mu=\frac{12}{7}mu$